Mixing
Prove that $n^7+7$ can never be a perfect square.
$begingroup$
Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
elementary-number-theory
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
$endgroup$
add a comment |
$begingroup$
Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
elementary-number-theory
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
$endgroup$
$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
$endgroup$
– JB King
Mar 22 '14 at 5:10
$begingroup$
How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
$endgroup$
– XYZT
Mar 22 '14 at 5:10
$begingroup$
Please see duplicate here at MSE. I just saw the same question may be from a different OP.
$endgroup$
– DeepSea
Mar 22 '14 at 5:13
$begingroup$
@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
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– John Marty
Mar 22 '14 at 8:11
$begingroup$
@Kf-Sansoo Where? Link?
$endgroup$
– mysatellite
Dec 30 '15 at 1:16
add a comment |
$begingroup$
Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
elementary-number-theory
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
$endgroup$
Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
elementary-number-theory
elementary-number-theory
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
asked Mar 22 '14 at 4:43
John MartyJohn Marty
2,0091136
2,0091136
$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
$endgroup$
– JB King
Mar 22 '14 at 5:10
$begingroup$
How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
$endgroup$
– XYZT
Mar 22 '14 at 5:10
$begingroup$
Please see duplicate here at MSE. I just saw the same question may be from a different OP.
$endgroup$
– DeepSea
Mar 22 '14 at 5:13
$begingroup$
@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
$endgroup$
– John Marty
Mar 22 '14 at 8:11
$begingroup$
@Kf-Sansoo Where? Link?
$endgroup$
– mysatellite
Dec 30 '15 at 1:16
add a comment |
$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
$endgroup$
– JB King
Mar 22 '14 at 5:10
$begingroup$
How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
$endgroup$
– XYZT
Mar 22 '14 at 5:10
$begingroup$
Please see duplicate here at MSE. I just saw the same question may be from a different OP.
$endgroup$
– DeepSea
Mar 22 '14 at 5:13
$begingroup$
@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
$endgroup$
– John Marty
Mar 22 '14 at 8:11
$begingroup$
@Kf-Sansoo Where? Link?
$endgroup$
– mysatellite
Dec 30 '15 at 1:16
$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
$endgroup$
– JB King
Mar 22 '14 at 5:10
$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
$endgroup$
– JB King
Mar 22 '14 at 5:10
$begingroup$
How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
$endgroup$
– XYZT
Mar 22 '14 at 5:10
$begingroup$
How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
$endgroup$
– XYZT
Mar 22 '14 at 5:10
$begingroup$
Please see duplicate here at MSE. I just saw the same question may be from a different OP.
$endgroup$
– DeepSea
Mar 22 '14 at 5:13
$begingroup$
Please see duplicate here at MSE. I just saw the same question may be from a different OP.
$endgroup$
– DeepSea
Mar 22 '14 at 5:13
$begingroup$
@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
$endgroup$
– John Marty
Mar 22 '14 at 8:11
$begingroup$
@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
$endgroup$
– John Marty
Mar 22 '14 at 8:11
$begingroup$
@Kf-Sansoo Where? Link?
$endgroup$
– mysatellite
Dec 30 '15 at 1:16
$begingroup$
@Kf-Sansoo Where? Link?
$endgroup$
– mysatellite
Dec 30 '15 at 1:16
add a comment |
3 Answers
3
active
oldest
votes
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To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$sum_{0 le k le 7} binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = sum_{1 le k le 7} binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m sum_{1 le k le 7} binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = mcdot M.$
Note that $M = m^6 -7cdot 2cdot m^5 + - + -21cdot 2^5m + 7cdot 2^6$
Hence $gcd(m,M)$ is a divisor of $7cdot 2^6$
Note that in $Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $gcd(m,M)$ is either $1$ or $7$.
Observe that in $Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $gcd(m,M)=1$
Now if $m cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $gcd(m,M) = 1$.
This implies that $m = 1 pmod 4$ and $n = m - 2 = - 1 . pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
$endgroup$
1
$begingroup$
BTW you can use mod and pmod.$7mod4$renders as $7mod4$ and$mequiv1pmod4$gives $mequiv1pmod4$.
$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
add a comment |
$begingroup$
Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$nequiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-n^5+n^4-n^3+n^2-n+1)=4b^2+11^2.$$
Now, notice $n+2equiv 3 (mod 4)$ and $n^6-n^5+n^4-n^3+n^2-n+1equiv 1 (mod 4)$, the left side of the above equation is $3 (mod 4)$, while the right hand size of the above equation is obviously $1 (mod 4)$. The contradiction completes the proof.
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
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add a comment |
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If you showed both of those things, then both must be true.
Can you show that squares cannot be 5 mod 8?
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$sum_{0 le k le 7} binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = sum_{1 le k le 7} binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m sum_{1 le k le 7} binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = mcdot M.$
Note that $M = m^6 -7cdot 2cdot m^5 + - + -21cdot 2^5m + 7cdot 2^6$
Hence $gcd(m,M)$ is a divisor of $7cdot 2^6$
Note that in $Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $gcd(m,M)$ is either $1$ or $7$.
Observe that in $Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $gcd(m,M)=1$
Now if $m cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $gcd(m,M) = 1$.
This implies that $m = 1 pmod 4$ and $n = m - 2 = - 1 . pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
$endgroup$
1
$begingroup$
BTW you can use mod and pmod.$7mod4$renders as $7mod4$ and$mequiv1pmod4$gives $mequiv1pmod4$.
$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
add a comment |
$begingroup$
To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$sum_{0 le k le 7} binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = sum_{1 le k le 7} binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m sum_{1 le k le 7} binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = mcdot M.$
Note that $M = m^6 -7cdot 2cdot m^5 + - + -21cdot 2^5m + 7cdot 2^6$
Hence $gcd(m,M)$ is a divisor of $7cdot 2^6$
Note that in $Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $gcd(m,M)$ is either $1$ or $7$.
Observe that in $Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $gcd(m,M)=1$
Now if $m cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $gcd(m,M) = 1$.
This implies that $m = 1 pmod 4$ and $n = m - 2 = - 1 . pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
$endgroup$
1
$begingroup$
BTW you can use mod and pmod.$7mod4$renders as $7mod4$ and$mequiv1pmod4$gives $mequiv1pmod4$.
$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
add a comment |
$begingroup$
To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$sum_{0 le k le 7} binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = sum_{1 le k le 7} binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m sum_{1 le k le 7} binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = mcdot M.$
Note that $M = m^6 -7cdot 2cdot m^5 + - + -21cdot 2^5m + 7cdot 2^6$
Hence $gcd(m,M)$ is a divisor of $7cdot 2^6$
Note that in $Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $gcd(m,M)$ is either $1$ or $7$.
Observe that in $Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $gcd(m,M)=1$
Now if $m cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $gcd(m,M) = 1$.
This implies that $m = 1 pmod 4$ and $n = m - 2 = - 1 . pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
$endgroup$
To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$sum_{0 le k le 7} binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = sum_{1 le k le 7} binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m sum_{1 le k le 7} binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = mcdot M.$
Note that $M = m^6 -7cdot 2cdot m^5 + - + -21cdot 2^5m + 7cdot 2^6$
Hence $gcd(m,M)$ is a divisor of $7cdot 2^6$
Note that in $Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $gcd(m,M)$ is either $1$ or $7$.
Observe that in $Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $gcd(m,M)=1$
Now if $m cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $gcd(m,M) = 1$.
This implies that $m = 1 pmod 4$ and $n = m - 2 = - 1 . pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
edited Dec 30 '15 at 15:46
mysatellite
2,14221231
2,14221231
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
answered Mar 22 '14 at 5:18
DeepSeaDeepSea
71.2k54487
71.2k54487
1
$begingroup$
BTW you can use mod and pmod.$7mod4$renders as $7mod4$ and$mequiv1pmod4$gives $mequiv1pmod4$.
$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
add a comment |
1
$begingroup$
BTW you can use mod and pmod.$7mod4$renders as $7mod4$ and$mequiv1pmod4$gives $mequiv1pmod4$.
$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
1
1
$begingroup$
BTW you can use mod and pmod.
$7mod4$ renders as $7mod4$ and $mequiv1pmod4$ gives $mequiv1pmod4$.$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
BTW you can use mod and pmod.
$7mod4$ renders as $7mod4$ and $mequiv1pmod4$ gives $mequiv1pmod4$.$endgroup$
– Martin Sleziak
Mar 22 '14 at 8:24
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
$begingroup$
@MartinSleziak: sure.
$endgroup$
– DeepSea
Mar 27 '14 at 17:51
add a comment |
$begingroup$
Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$nequiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-n^5+n^4-n^3+n^2-n+1)=4b^2+11^2.$$
Now, notice $n+2equiv 3 (mod 4)$ and $n^6-n^5+n^4-n^3+n^2-n+1equiv 1 (mod 4)$, the left side of the above equation is $3 (mod 4)$, while the right hand size of the above equation is obviously $1 (mod 4)$. The contradiction completes the proof.
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
add a comment |
$begingroup$
Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$nequiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-n^5+n^4-n^3+n^2-n+1)=4b^2+11^2.$$
Now, notice $n+2equiv 3 (mod 4)$ and $n^6-n^5+n^4-n^3+n^2-n+1equiv 1 (mod 4)$, the left side of the above equation is $3 (mod 4)$, while the right hand size of the above equation is obviously $1 (mod 4)$. The contradiction completes the proof.
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
add a comment |
$begingroup$
Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$nequiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-n^5+n^4-n^3+n^2-n+1)=4b^2+11^2.$$
Now, notice $n+2equiv 3 (mod 4)$ and $n^6-n^5+n^4-n^3+n^2-n+1equiv 1 (mod 4)$, the left side of the above equation is $3 (mod 4)$, while the right hand size of the above equation is obviously $1 (mod 4)$. The contradiction completes the proof.
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
$endgroup$
Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$nequiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-n^5+n^4-n^3+n^2-n+1)=4b^2+11^2.$$
Now, notice $n+2equiv 3 (mod 4)$ and $n^6-n^5+n^4-n^3+n^2-n+1equiv 1 (mod 4)$, the left side of the above equation is $3 (mod 4)$, while the right hand size of the above equation is obviously $1 (mod 4)$. The contradiction completes the proof.
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
answered Jan 14 at 4:11
Xiaohai ZhangXiaohai Zhang
212
212
add a comment |
add a comment |
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If you showed both of those things, then both must be true.
Can you show that squares cannot be 5 mod 8?
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
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$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
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– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
add a comment |
$begingroup$
If you showed both of those things, then both must be true.
Can you show that squares cannot be 5 mod 8?
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
$endgroup$
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
add a comment |
$begingroup$
If you showed both of those things, then both must be true.
Can you show that squares cannot be 5 mod 8?
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
$endgroup$
If you showed both of those things, then both must be true.
Can you show that squares cannot be 5 mod 8?
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
answered Mar 22 '14 at 4:47
Empy2Empy2
33.5k12261
33.5k12261
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
add a comment |
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
I think you misread the question. $n$ doen't have to be a perfect square, $n^7+7$ does.
$endgroup$
– John Marty
Mar 22 '14 at 4:51
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
$begingroup$
Ah. Whoops. Yes, I misread it.
$endgroup$
– Empy2
Mar 22 '14 at 6:45
add a comment |
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$begingroup$
If n is even, then $n^7$ is also going to be even which isn't the same as what you showed, thus you missed something there.
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– JB King
Mar 22 '14 at 5:10
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How did you show that $n equiv 5 pmod{8}$ or $n equiv 9 pmod{16}$?
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– XYZT
Mar 22 '14 at 5:10
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Please see duplicate here at MSE. I just saw the same question may be from a different OP.
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– DeepSea
Mar 22 '14 at 5:13
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@ Nikihil by checking values of n mod 8 and then applying the fact that squares can only be congruent to 0,1,4 mod 8
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– John Marty
Mar 22 '14 at 8:11
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@Kf-Sansoo Where? Link?
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– mysatellite
Dec 30 '15 at 1:16