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If $2n+1$ and $3n+1$ are perfect squares, then prove that $8|n$.
If for some number $nin mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder
discrete-mathematics modular-arithmetic divisibility diophantine-equations pell-type-equations
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
add a comment |
If for some number $nin mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder
discrete-mathematics modular-arithmetic divisibility diophantine-equations pell-type-equations
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
add a comment |
If for some number $nin mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder
discrete-mathematics modular-arithmetic divisibility diophantine-equations pell-type-equations
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
If for some number $nin mathbb N$, the numbers $2n+1$ and $3n+1$ are perfect squares of integers, then prove that $8|n$.
if $2n+1=m^2$ and $3n+1=k^2$ then $k^2-m^2=3n-2n+1-1=n$ now I need to show that $8|k^2-m^2$ when you divide a some number $m^2$ with $8$ then remainder is $0,1,4$, so I need to show that $m^2$ and $k^2$ have the same remainder. But I do not know is this good way because I do not know how to prove that they must have the same remainder
discrete-mathematics modular-arithmetic divisibility diophantine-equations pell-type-equations
discrete-mathematics modular-arithmetic divisibility diophantine-equations pell-type-equations
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
edited Nov 21 '18 at 18:39
Batominovski
33.9k33294
33.9k33294
asked Nov 21 '18 at 16:47
Marko Škorić
70310
asked Nov 21 '18 at 16:47
Marko Škorić
70310
asked Nov 21 '18 at 16:47
Marko Škorić
70310
70310
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
If $k$ is odd, then $k^2equiv1 mod 8$. Hence $3n+1equiv1mod 8$, $2n+1equiv 1mod 8$, so
$$(3n+1)-(2n+1)equiv 1-1equiv 0mod 8$$
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
add a comment |
If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.
Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.
In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is,
$$(1+sqrt{-2})(1-sqrt{-2})a^2=1+2b^2=(1+sqrt{-2}b)(1-sqrt{-2}b).$$
Note that $Bbb{Q}(sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $gcd(1+sqrt{-2}b,1-sqrt{-2}b)=1$, we get that either
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
or $$frac{1-sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
for some $u,vinBbb{Z}$. But up to sign switching $bto -b$, we can assume that
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2=u^2-2v^2+2uvsqrt{-2}.$$
That is,
$$1+sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)sqrt{-2}.$$
So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation
$$x^2-6y^2=1.$$
The solutions are known $$x+sqrt{6}y=pm(5+2sqrt{6})^t$$
where $tinBbb{Z}$. Since the sign switching $(u,v)to(-u,-v)$ does not change anything, we can assume that $$u-2v+sqrt{6}v=(5+2sqrt{6})^t.$$
So, $$u-2v=frac{(5+2sqrt{6})^t+(5-2sqrt{6})^{t}}{2}$$
and $$v=frac{(5+2sqrt{6})^t-(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$u=frac{(2+sqrt{6})(5+2sqrt{6})^t-(2-sqrt{6})(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$b=u^2-2v^2+2uv=frac{(2+sqrt{6})(5+2sqrt{6})^{2t}+(2-sqrt{6})(5-2sqrt{6})^{2t}}{4}.$$
This gives
$$a=sqrt{frac{1+2b^2}{3}}=u^2+2v^2=frac{(3+sqrt{6})(5+2sqrt{6})^{2t}+(3-sqrt{6})(5-2sqrt{6})^{2t}}{6}.$$
So, we have
$$n=frac{(5+2sqrt{6})^{4t+1}-10+(5-2sqrt{6})^{4t+1}}{24},$$
where $tinmathbb{Z}$. So the first seven values of $n$ are
$$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$
That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with
$$n(t-1)+n(t+1)=9602 n(t)+4000$$
for all $tinmathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.
I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=frac{(5+2sqrt{6})^{2t+1}-10+(5-2sqrt{6})^{2t+1}}{24},$$
for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$
We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and
$$a_{t+2}=10a_{t+1}-a_t,$$
as well as $b_0=1$, $b_1=11$, and
$$b_{t+2}=10b_{t+1}-b_t.$$
$$
begin{array}{ |c|c|c|c| }
hline
t & n_t & a_t & b_t \ hline
0 & 0 & 1 &1 \
1 & 40 & 9& 11 \
2 & 3960 & 89 & 109 \
3 & 388080 & 881 &1079\
4 & 38027920 & 8721 & 10681 \
5 & 3726348120 & 86329 & 105731 \
6 & 365144087880 & 854569 & 1046629
\hline
end{array}
$$
answered Nov 21 '18 at 16:56
Snookie
1,30517
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
|
show 1 more comment
It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4not mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4not mid b-a$ and $4mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
add a comment |
Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8mid n$.
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
add a comment |
We have $2n+1, 3n+1 equiv {0,1,4 } mod 8,$ whereupon checking $n = 0, 1, dots, 7,$ we see that only $n equiv 0 mod 8$ works.
Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 equiv {0, 1} mod 5$ implies $n equiv 0 mod 5$ by checking $0, 1, dots, 4.$
Since $n=40$ works, this is the strongest result that we can prove.
answered Nov 21 '18 at 20:53
Display name
811314
add a comment |
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5 Answers
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5 Answers
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If $k$ is odd, then $k^2equiv1 mod 8$. Hence $3n+1equiv1mod 8$, $2n+1equiv 1mod 8$, so
$$(3n+1)-(2n+1)equiv 1-1equiv 0mod 8$$
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
add a comment |
If $k$ is odd, then $k^2equiv1 mod 8$. Hence $3n+1equiv1mod 8$, $2n+1equiv 1mod 8$, so
$$(3n+1)-(2n+1)equiv 1-1equiv 0mod 8$$
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
add a comment |
If $k$ is odd, then $k^2equiv1 mod 8$. Hence $3n+1equiv1mod 8$, $2n+1equiv 1mod 8$, so
$$(3n+1)-(2n+1)equiv 1-1equiv 0mod 8$$
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
If $k$ is odd, then $k^2equiv1 mod 8$. Hence $3n+1equiv1mod 8$, $2n+1equiv 1mod 8$, so
$$(3n+1)-(2n+1)equiv 1-1equiv 0mod 8$$
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
answered Nov 21 '18 at 17:00
Nicolas FRANCOIS
3,6221516
3,6221516
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
add a comment |
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
You need to prove that $k$ is odd to apply that, but you have not done so.
– Bill Dubuque
Nov 21 '18 at 20:32
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Minor issue. How do we know $3n+1$ and $2n + 1$ are odd?
– fleablood
Nov 21 '18 at 20:49
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
Damned, you're right :-( Didn't see this one coming :-)
– Nicolas FRANCOIS
Nov 22 '18 at 20:32
add a comment |
If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.
Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.
In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is,
$$(1+sqrt{-2})(1-sqrt{-2})a^2=1+2b^2=(1+sqrt{-2}b)(1-sqrt{-2}b).$$
Note that $Bbb{Q}(sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $gcd(1+sqrt{-2}b,1-sqrt{-2}b)=1$, we get that either
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
or $$frac{1-sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
for some $u,vinBbb{Z}$. But up to sign switching $bto -b$, we can assume that
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2=u^2-2v^2+2uvsqrt{-2}.$$
That is,
$$1+sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)sqrt{-2}.$$
So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation
$$x^2-6y^2=1.$$
The solutions are known $$x+sqrt{6}y=pm(5+2sqrt{6})^t$$
where $tinBbb{Z}$. Since the sign switching $(u,v)to(-u,-v)$ does not change anything, we can assume that $$u-2v+sqrt{6}v=(5+2sqrt{6})^t.$$
So, $$u-2v=frac{(5+2sqrt{6})^t+(5-2sqrt{6})^{t}}{2}$$
and $$v=frac{(5+2sqrt{6})^t-(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$u=frac{(2+sqrt{6})(5+2sqrt{6})^t-(2-sqrt{6})(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$b=u^2-2v^2+2uv=frac{(2+sqrt{6})(5+2sqrt{6})^{2t}+(2-sqrt{6})(5-2sqrt{6})^{2t}}{4}.$$
This gives
$$a=sqrt{frac{1+2b^2}{3}}=u^2+2v^2=frac{(3+sqrt{6})(5+2sqrt{6})^{2t}+(3-sqrt{6})(5-2sqrt{6})^{2t}}{6}.$$
So, we have
$$n=frac{(5+2sqrt{6})^{4t+1}-10+(5-2sqrt{6})^{4t+1}}{24},$$
where $tinmathbb{Z}$. So the first seven values of $n$ are
$$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$
That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with
$$n(t-1)+n(t+1)=9602 n(t)+4000$$
for all $tinmathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.
I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=frac{(5+2sqrt{6})^{2t+1}-10+(5-2sqrt{6})^{2t+1}}{24},$$
for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$
We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and
$$a_{t+2}=10a_{t+1}-a_t,$$
as well as $b_0=1$, $b_1=11$, and
$$b_{t+2}=10b_{t+1}-b_t.$$
$$
begin{array}{ |c|c|c|c| }
hline
t & n_t & a_t & b_t \ hline
0 & 0 & 1 &1 \
1 & 40 & 9& 11 \
2 & 3960 & 89 & 109 \
3 & 388080 & 881 &1079\
4 & 38027920 & 8721 & 10681 \
5 & 3726348120 & 86329 & 105731 \
6 & 365144087880 & 854569 & 1046629
\hline
end{array}
$$
answered Nov 21 '18 at 16:56
Snookie
1,30517
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
|
show 1 more comment
If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.
Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.
In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is,
$$(1+sqrt{-2})(1-sqrt{-2})a^2=1+2b^2=(1+sqrt{-2}b)(1-sqrt{-2}b).$$
Note that $Bbb{Q}(sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $gcd(1+sqrt{-2}b,1-sqrt{-2}b)=1$, we get that either
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
or $$frac{1-sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
for some $u,vinBbb{Z}$. But up to sign switching $bto -b$, we can assume that
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2=u^2-2v^2+2uvsqrt{-2}.$$
That is,
$$1+sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)sqrt{-2}.$$
So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation
$$x^2-6y^2=1.$$
The solutions are known $$x+sqrt{6}y=pm(5+2sqrt{6})^t$$
where $tinBbb{Z}$. Since the sign switching $(u,v)to(-u,-v)$ does not change anything, we can assume that $$u-2v+sqrt{6}v=(5+2sqrt{6})^t.$$
So, $$u-2v=frac{(5+2sqrt{6})^t+(5-2sqrt{6})^{t}}{2}$$
and $$v=frac{(5+2sqrt{6})^t-(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$u=frac{(2+sqrt{6})(5+2sqrt{6})^t-(2-sqrt{6})(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$b=u^2-2v^2+2uv=frac{(2+sqrt{6})(5+2sqrt{6})^{2t}+(2-sqrt{6})(5-2sqrt{6})^{2t}}{4}.$$
This gives
$$a=sqrt{frac{1+2b^2}{3}}=u^2+2v^2=frac{(3+sqrt{6})(5+2sqrt{6})^{2t}+(3-sqrt{6})(5-2sqrt{6})^{2t}}{6}.$$
So, we have
$$n=frac{(5+2sqrt{6})^{4t+1}-10+(5-2sqrt{6})^{4t+1}}{24},$$
where $tinmathbb{Z}$. So the first seven values of $n$ are
$$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$
That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with
$$n(t-1)+n(t+1)=9602 n(t)+4000$$
for all $tinmathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.
I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=frac{(5+2sqrt{6})^{2t+1}-10+(5-2sqrt{6})^{2t+1}}{24},$$
for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$
We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and
$$a_{t+2}=10a_{t+1}-a_t,$$
as well as $b_0=1$, $b_1=11$, and
$$b_{t+2}=10b_{t+1}-b_t.$$
$$
begin{array}{ |c|c|c|c| }
hline
t & n_t & a_t & b_t \ hline
0 & 0 & 1 &1 \
1 & 40 & 9& 11 \
2 & 3960 & 89 & 109 \
3 & 388080 & 881 &1079\
4 & 38027920 & 8721 & 10681 \
5 & 3726348120 & 86329 & 105731 \
6 & 365144087880 & 854569 & 1046629
\hline
end{array}
$$
answered Nov 21 '18 at 16:56
Snookie
1,30517
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
|
show 1 more comment
If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.
Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.
In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is,
$$(1+sqrt{-2})(1-sqrt{-2})a^2=1+2b^2=(1+sqrt{-2}b)(1-sqrt{-2}b).$$
Note that $Bbb{Q}(sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $gcd(1+sqrt{-2}b,1-sqrt{-2}b)=1$, we get that either
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
or $$frac{1-sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
for some $u,vinBbb{Z}$. But up to sign switching $bto -b$, we can assume that
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2=u^2-2v^2+2uvsqrt{-2}.$$
That is,
$$1+sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)sqrt{-2}.$$
So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation
$$x^2-6y^2=1.$$
The solutions are known $$x+sqrt{6}y=pm(5+2sqrt{6})^t$$
where $tinBbb{Z}$. Since the sign switching $(u,v)to(-u,-v)$ does not change anything, we can assume that $$u-2v+sqrt{6}v=(5+2sqrt{6})^t.$$
So, $$u-2v=frac{(5+2sqrt{6})^t+(5-2sqrt{6})^{t}}{2}$$
and $$v=frac{(5+2sqrt{6})^t-(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$u=frac{(2+sqrt{6})(5+2sqrt{6})^t-(2-sqrt{6})(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$b=u^2-2v^2+2uv=frac{(2+sqrt{6})(5+2sqrt{6})^{2t}+(2-sqrt{6})(5-2sqrt{6})^{2t}}{4}.$$
This gives
$$a=sqrt{frac{1+2b^2}{3}}=u^2+2v^2=frac{(3+sqrt{6})(5+2sqrt{6})^{2t}+(3-sqrt{6})(5-2sqrt{6})^{2t}}{6}.$$
So, we have
$$n=frac{(5+2sqrt{6})^{4t+1}-10+(5-2sqrt{6})^{4t+1}}{24},$$
where $tinmathbb{Z}$. So the first seven values of $n$ are
$$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$
That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with
$$n(t-1)+n(t+1)=9602 n(t)+4000$$
for all $tinmathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.
I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=frac{(5+2sqrt{6})^{2t+1}-10+(5-2sqrt{6})^{2t+1}}{24},$$
for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$
We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and
$$a_{t+2}=10a_{t+1}-a_t,$$
as well as $b_0=1$, $b_1=11$, and
$$b_{t+2}=10b_{t+1}-b_t.$$
$$
begin{array}{ |c|c|c|c| }
hline
t & n_t & a_t & b_t \ hline
0 & 0 & 1 &1 \
1 & 40 & 9& 11 \
2 & 3960 & 89 & 109 \
3 & 388080 & 881 &1079\
4 & 38027920 & 8721 & 10681 \
5 & 3726348120 & 86329 & 105731 \
6 & 365144087880 & 854569 & 1046629
\hline
end{array}
$$
answered Nov 21 '18 at 16:56
Snookie
1,30517
If $2n+1=a^2$ and $3n+1=b^2$, then $a$ is odd. So $a=2k+1$ and $2n+1=(2k+1)^2=4k^2+4k+1$. Thus, $n=2k^2+2k=2k(k+1)$. This shows that $n$ is divisible by $4$ because $k(k+1)$ is even.
Now, $b^2=3n+1=6k(k+1)+1$. So, $b$ is also odd and $b=2l+1$. Then, $(2l+1)^2=6k(k+1)+1$ implies $2l(l+1)=3k(k+1)$. That is $k(k+1)$ is divisible by $4$ because $l(l+1)$ is even. Because $n=2k(k+1)$ and $k(k+1)$ is divisible by $4$, $n$ is divisible by $8$.
In fact, $3a^2-2b^2=3(2n+1)-2(3n+1)=1$. That is,
$$(1+sqrt{-2})(1-sqrt{-2})a^2=1+2b^2=(1+sqrt{-2}b)(1-sqrt{-2}b).$$
Note that $Bbb{Q}(sqrt{-2})$ is a quadratic field with class number $1$, it is a ufd and we can talk about gcd. Because $gcd(1+sqrt{-2}b,1-sqrt{-2}b)=1$, we get that either
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
or $$frac{1-sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2$$
for some $u,vinBbb{Z}$. But up to sign switching $bto -b$, we can assume that
$$frac{1+sqrt{-2}b}{1+sqrt{-2}}=(u+sqrt{-2}v)^2=u^2-2v^2+2uvsqrt{-2}.$$
That is,
$$1+sqrt{-2}b=u^2-2v^2-4uv+(u^2-2v^2+2uv)sqrt{-2}.$$
So, $u^2-2v^2-4uv=1$ and so $(u-2v)^2-6v^2=1$. So, $(x,y)=(u-2v,v)$ is a solution to the Pell-type equation
$$x^2-6y^2=1.$$
The solutions are known $$x+sqrt{6}y=pm(5+2sqrt{6})^t$$
where $tinBbb{Z}$. Since the sign switching $(u,v)to(-u,-v)$ does not change anything, we can assume that $$u-2v+sqrt{6}v=(5+2sqrt{6})^t.$$
So, $$u-2v=frac{(5+2sqrt{6})^t+(5-2sqrt{6})^{t}}{2}$$
and $$v=frac{(5+2sqrt{6})^t-(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$u=frac{(2+sqrt{6})(5+2sqrt{6})^t-(2-sqrt{6})(5-2sqrt{6})^{t}}{2sqrt{6}}.$$
That is,
$$b=u^2-2v^2+2uv=frac{(2+sqrt{6})(5+2sqrt{6})^{2t}+(2-sqrt{6})(5-2sqrt{6})^{2t}}{4}.$$
This gives
$$a=sqrt{frac{1+2b^2}{3}}=u^2+2v^2=frac{(3+sqrt{6})(5+2sqrt{6})^{2t}+(3-sqrt{6})(5-2sqrt{6})^{2t}}{6}.$$
So, we have
$$n=frac{(5+2sqrt{6})^{4t+1}-10+(5-2sqrt{6})^{4t+1}}{24},$$
where $tinmathbb{Z}$. So the first seven values of $n$ are
$$n=0,40,3960, 388080,38027920,3726348120,365144087880.$$
That is, $n=n(t)$ satisfies $n(0)=0$ and $n(-1)=40$ with
$$n(t-1)+n(t+1)=9602 n(t)+4000$$
for all $tinmathbb{Z}$. So, not only $8$ divides $n$, $40$ divides $n$ for every such $n$.
I think it is easier to re-parametrize $n$ using non-negative integers instead of all integers. Let $$n_t=frac{(5+2sqrt{6})^{2t+1}-10+(5-2sqrt{6})^{2t+1}}{24},$$
for non-negative integer $t$. So, $n_0=0$, $n_1=40$, and $$n_{t+2}=98n_{t+1}-n_t+40.$$
We have the corresponding $a=a_t$ and $b=b_t$ to $n=n_t$: $a_0=1$, $a_1=9$, and
$$a_{t+2}=10a_{t+1}-a_t,$$
as well as $b_0=1$, $b_1=11$, and
$$b_{t+2}=10b_{t+1}-b_t.$$
$$
begin{array}{ |c|c|c|c| }
hline
t & n_t & a_t & b_t \ hline
0 & 0 & 1 &1 \
1 & 40 & 9& 11 \
2 & 3960 & 89 & 109 \
3 & 388080 & 881 &1079\
4 & 38027920 & 8721 & 10681 \
5 & 3726348120 & 86329 & 105731 \
6 & 365144087880 & 854569 & 1046629
\hline
end{array}
$$
answered Nov 21 '18 at 16:56
Snookie
1,30517
edited Nov 22 '18 at 6:49
answered Nov 21 '18 at 16:56
Snookie
1,30517
answered Nov 21 '18 at 16:56
Snookie
1,30517
answered Nov 21 '18 at 16:56
Snookie
1,30517
1,30517
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
|
show 1 more comment
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
Does $4|n$ as well?
– Yadati Kiran
Nov 21 '18 at 16:59
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
If $8mid n$, then $4mid n$.
– Snookie
Nov 21 '18 at 17:00
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
I don't know what you did, but your answer was off by a factor of $2$. So, I fixed it.
– Batominovski
Nov 21 '18 at 18:31
1
1
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
I have a feeling that you can directly apply some results from Pell's equations, without having to deal with $mathbb{Q}(sqrt{-2})$. But I haven't thought about it well enough.
– Batominovski
Nov 21 '18 at 18:41
2
2
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
@Displayname I don't think Snookie was trying to prove $40mid n$. The aim was probably to find all $n$. The result $40mid n$ is a nice little consequence. And I think we can use Pell's equation $x^2-6y^2=-2$ or $x^2-6y^2=3$ without having to deal with $mathbb{Q}(sqrt{-2})$. However, I am not sure whether these Pell-type equations have unique fundamental solutions.
– Batominovski
Nov 21 '18 at 21:09
|
show 1 more comment
It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4not mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4not mid b-a$ and $4mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
add a comment |
It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4not mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4not mid b-a$ and $4mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
add a comment |
It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4not mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4not mid b-a$ and $4mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
It's a standard exercise that if $b > a$ and $a$ and $b$ are both odd that
i) $a + b$ and $a-b$ are both even and that
ii) exactly one of $a+b$ and $a-b$ is divisible by $4$.
Pf: Let $b = 2m + 1$ and $a = 2n+1$ so $b-a = 2(m-n)$ and $b+a = 2(m+n+1)$ are both even.
If $m$ and $n$ are both even or both odd then $m-n$ is even and $m+n + 1$ is odd so $4|b-a$ but $4not mid b+a$. If $m$ and $n$ are opposite paritty then the exact opposite occurs; $m-n$ is odd and $m+n +1$ is even so $4not mid b-a$ and $4mid b+a$.
As a consequence $b^2 - a^2 = (b-a)(b+a)$ is divisible by $8$.
So if we have $2n + 1 = a^2$ then $a^2$ is odd so $a$ is odd.
AND we have $2n = a^2 -1 = (a-1)(a+1)$ is divisible by $8$ so $n$ is even.
And if $3n + 1=b^2$ then we have $n$ is even so $b^2$ is odd so $b$ is odd.
So $n = (3n+1)-(2n+1) =b^2 - a^2$ where $a,b$ are both odd..... which is divisible by $8$.
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
answered Nov 21 '18 at 20:05
fleablood
68.4k22685
68.4k22685
add a comment |
add a comment |
Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8mid n$.
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
add a comment |
Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8mid n$.
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
add a comment |
Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8mid n$.
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
Odd squares are all congruent to $1$ mod $8$, so if $2n+1$ is a square, we must have $4mid n$. Writing $n=4n'$, we now have $3n+1=12n'+1$, which, if it's square, must also be congruent to $1$ mod $8$, so we must have $2mid n'$. Writing $n'=2n''$, we have $n=4n'=8n''$, so $8mid n$.
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
answered Nov 21 '18 at 20:44
Barry Cipra
59.3k653125
59.3k653125
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We have $2n+1, 3n+1 equiv {0,1,4 } mod 8,$ whereupon checking $n = 0, 1, dots, 7,$ we see that only $n equiv 0 mod 8$ works.
Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 equiv {0, 1} mod 5$ implies $n equiv 0 mod 5$ by checking $0, 1, dots, 4.$
Since $n=40$ works, this is the strongest result that we can prove.
answered Nov 21 '18 at 20:53
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811314
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We have $2n+1, 3n+1 equiv {0,1,4 } mod 8,$ whereupon checking $n = 0, 1, dots, 7,$ we see that only $n equiv 0 mod 8$ works.
Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 equiv {0, 1} mod 5$ implies $n equiv 0 mod 5$ by checking $0, 1, dots, 4.$
Since $n=40$ works, this is the strongest result that we can prove.
answered Nov 21 '18 at 20:53
Display name
811314
add a comment |
We have $2n+1, 3n+1 equiv {0,1,4 } mod 8,$ whereupon checking $n = 0, 1, dots, 7,$ we see that only $n equiv 0 mod 8$ works.
Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 equiv {0, 1} mod 5$ implies $n equiv 0 mod 5$ by checking $0, 1, dots, 4.$
Since $n=40$ works, this is the strongest result that we can prove.
answered Nov 21 '18 at 20:53
Display name
811314
We have $2n+1, 3n+1 equiv {0,1,4 } mod 8,$ whereupon checking $n = 0, 1, dots, 7,$ we see that only $n equiv 0 mod 8$ works.
Snookie has shown that $40 | n.$ Indeed, we can show $5 | n$ in a similar fashion by noting that $2n + 1 equiv {0, 1} mod 5$ implies $n equiv 0 mod 5$ by checking $0, 1, dots, 4.$
Since $n=40$ works, this is the strongest result that we can prove.
answered Nov 21 '18 at 20:53
Display name
811314
edited Nov 21 '18 at 21:05
answered Nov 21 '18 at 20:53
Display name
811314
answered Nov 21 '18 at 20:53
Display name
811314
answered Nov 21 '18 at 20:53
Display name
811314
811314
add a comment |
add a comment |
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