# Math Help - Diophantine equation

1. ## Diophantine equation

How to solve the following Diophantine equation?
$a^2+b^2+c^2=a^2b^2$

2. Originally Posted by doug
How to solve the following Diophantine equation?
$a^2+b^2+c^2=a^2b^2$
That equation has no integer solutions. Remember that $x^2$ is congruent to 0 (mod 4) if $x$ is even, and congruent t0 1 (mod 4) if $x$ is odd.

So $a^2b^2\equiv0\pmod4$ unless $a$ and $b$ are both odd. But if $a$ and $b$ are both odd the $a^2+b^2+c^2$ will be congruent to 2 or 3 (mod 4), so that case cannot arise.

Therefore $a^2b^2\equiv0\pmod4$ and hence $a^2+b^2+c^2\equiv0\pmod4$. But that implies that $a$, $b$ and $c$ are all even, say $a=2a'$, $b=2b'$, $c=2c'$. Then the equation becomes $a'^{\,2}+b'^{\,2}+c'^2=4a'^{\,2}b'^{\,2}$. But this time, the right side is obviously congruent to 0 (mod 4), so $a'$, $b'$ and $c'$ must all be even.

That argument can be continued indefinitely, so by Fermat's method of infinite descent it follows that there are no solutions.

3. Originally Posted by Opalg
That equation has no integer solutions. Remember that $x^2$ is congruent to 0 (mod 4) if $x$ is even, and congruent t0 1 (mod 4) if $x$ is odd.

So $a^2b^2\equiv0\pmod4$ unless $a$ and $b$ are both odd. But if $a$ and $b$ are both odd the $a^2+b^2+c^2$ will be congruent to 2 or 3 (mod 4), so that case cannot arise.

Therefore $a^2b^2\equiv0\pmod4$ and hence $a^2+b^2+c^2\equiv0\pmod4$. But that implies that $a$, $b$ and $c$ are all even, say $a=2a'$, $b=2b'$, $c=2c'$. Then the equation becomes $a'^{\,2}+b'^{\,2}+c'^2=4a'^{\,2}b'^{\,2}$. But this time, the right side is obviously congruent to 0 (mod 4), so $a'$, $b'$ and $c'$ must all be even.

That argument can be continued indefinitely, so by Fermat's method of infinite descent it follows that there are no solutions.

But the trivial one (0,0,0).

Tonio