Mixing
If $a^2+b^2+c^2=D$ and a and b are consecutive positive integers and $ ab = c $ then prove that $ D^{1/2} $...
$begingroup$
Here is my approach:
since a and b are consecutive integers there are two possibilities
$a = b+1$
and $a = b-1$
substituting this into the first equation we get
$ 2 b^2 + 2 b + c^2 + 1 = D $ after this I do not know what to do.
help in any form would be appreciated.
number-theory factoring
edited Jan 29 at 13:50
freehumorist
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
$endgroup$
add a comment |
$begingroup$
Here is my approach:
since a and b are consecutive integers there are two possibilities
$a = b+1$
and $a = b-1$
substituting this into the first equation we get
$ 2 b^2 + 2 b + c^2 + 1 = D $ after this I do not know what to do.
help in any form would be appreciated.
number-theory factoring
edited Jan 29 at 13:50
freehumorist
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
$endgroup$
1
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
1
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45
add a comment |
$begingroup$
Here is my approach:
since a and b are consecutive integers there are two possibilities
$a = b+1$
and $a = b-1$
substituting this into the first equation we get
$ 2 b^2 + 2 b + c^2 + 1 = D $ after this I do not know what to do.
help in any form would be appreciated.
number-theory factoring
edited Jan 29 at 13:50
freehumorist
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
$endgroup$
Here is my approach:
since a and b are consecutive integers there are two possibilities
$a = b+1$
and $a = b-1$
substituting this into the first equation we get
$ 2 b^2 + 2 b + c^2 + 1 = D $ after this I do not know what to do.
help in any form would be appreciated.
number-theory factoring
number-theory factoring
edited Jan 29 at 13:50
freehumorist
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
edited Jan 29 at 13:50
freehumorist
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
edited Jan 29 at 13:50
freehumorist
351214
edited Jan 29 at 13:50
freehumorist
351214
edited Jan 29 at 13:50
freehumorist
351214
351214
asked Jan 29 at 13:40
Mary TomMary Tom
326
asked Jan 29 at 13:40
Mary TomMary Tom
326
asked Jan 29 at 13:40
Mary TomMary Tom
326
326
1
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
1
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45
add a comment |
1
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
1
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45
1
1
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
1
1
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
$endgroup$
add a comment |
$begingroup$
here is how you can do it:
you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $
now use $ab=c$
and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing)
to get $b(b+1)=c$ or $b(b-1)=c$
plug this value of c to get
$c+c^2+1=D$,
$(c+1)^2=D$,
so $D^{1/2} = c+1$
and since the product of consecutive integers ($a$ and $b$) is always even
it means $c$ is even so $c+1$ is odd
edited Jan 29 at 13:51
freehumorist
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
$endgroup$
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
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active
oldest
votes
$begingroup$
Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
$endgroup$
add a comment |
$begingroup$
Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
$endgroup$
add a comment |
$begingroup$
Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
$endgroup$
Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
answered Jan 29 at 13:46
Michael LeeMichael Lee
4,8281930
4,8281930
add a comment |
add a comment |
$begingroup$
here is how you can do it:
you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $
now use $ab=c$
and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing)
to get $b(b+1)=c$ or $b(b-1)=c$
plug this value of c to get
$c+c^2+1=D$,
$(c+1)^2=D$,
so $D^{1/2} = c+1$
and since the product of consecutive integers ($a$ and $b$) is always even
it means $c$ is even so $c+1$ is odd
edited Jan 29 at 13:51
freehumorist
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
$endgroup$
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
add a comment |
$begingroup$
here is how you can do it:
you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $
now use $ab=c$
and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing)
to get $b(b+1)=c$ or $b(b-1)=c$
plug this value of c to get
$c+c^2+1=D$,
$(c+1)^2=D$,
so $D^{1/2} = c+1$
and since the product of consecutive integers ($a$ and $b$) is always even
it means $c$ is even so $c+1$ is odd
edited Jan 29 at 13:51
freehumorist
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
$endgroup$
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
add a comment |
$begingroup$
here is how you can do it:
you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $
now use $ab=c$
and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing)
to get $b(b+1)=c$ or $b(b-1)=c$
plug this value of c to get
$c+c^2+1=D$,
$(c+1)^2=D$,
so $D^{1/2} = c+1$
and since the product of consecutive integers ($a$ and $b$) is always even
it means $c$ is even so $c+1$ is odd
edited Jan 29 at 13:51
freehumorist
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
$endgroup$
here is how you can do it:
you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $
now use $ab=c$
and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing)
to get $b(b+1)=c$ or $b(b-1)=c$
plug this value of c to get
$c+c^2+1=D$,
$(c+1)^2=D$,
so $D^{1/2} = c+1$
and since the product of consecutive integers ($a$ and $b$) is always even
it means $c$ is even so $c+1$ is odd
edited Jan 29 at 13:51
freehumorist
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
edited Jan 29 at 13:51
freehumorist
351214
edited Jan 29 at 13:51
freehumorist
351214
edited Jan 29 at 13:51
freehumorist
351214
351214
answered Jan 29 at 13:47
John TomJohn Tom
858
answered Jan 29 at 13:47
John TomJohn Tom
858
answered Jan 29 at 13:47
John TomJohn Tom
858
858
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
add a comment |
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
$begingroup$
sorry for not formatting this using mathjax as i am using my friends laptop
$endgroup$
– John Tom
Jan 29 at 13:50
add a comment |
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1
$begingroup$
You haven't used the fact that $c=ab$.
$endgroup$
– lulu
Jan 29 at 13:43
1
$begingroup$
The two cases $a = b+1$ and $a = b-1$ are basically the same. One of $a$ and $b$ is $1$ larger than the other, and one of $a$ and $b$ happens to have the name $a$. You may safely assume the two properties coincide in the same variable. We call this "assuming without loss of generality".
$endgroup$
– Arthur
Jan 29 at 13:45