Find all integer solutions, $(ax-b)^2+(bx-a)^2=x$

Question

Find all $(,,) \in \mathbb{z}^3$ satisfying

$(−)^2+(−)^2=$.

I found $(0,0,0)$ and $(±1,±1,2)$.

And for the case $≠$ , the $\Delta$ of quadratic in $x$ is $$(4+1)^2−(2^2+2^2)^2=^2$$ we set $\Delta=^2$ so that $x$ may be an integer.

Equaling the general formula for pythagorean triples $(^2+^2,2,^2−^2)$ and after some confusing steps of substitutions involving $$, $$ and $$ found that $$2(+)^2+1=0$$

which implies no solutions.

But I'm afraid that my solution is wrong because I've never seen Pythagorean triples used like that.

Can someone give another solution maybe more clarifying?

Please edit your posting to show the work leading to the conclusion that $$a = b \implies 2(a+b)^2 + 1 = 0.$$ — user2661923, Jun 05 '22 at 18:58

score 0 · Accepted Answer · answered Jun 05 '22 at 15:37

0

If your computation is correct for the discriminant, then you can observe that: $$k^2 = (4ab+1+2a^2+2b^2)(4ab+1-2a^2-2b^2) = \left(2(a+b)^2+1\right)\left(1-2(a-b)^2\right) < 0$$ if $a\neq b.$ This forces $a = b$ in particular. Your method of using the general formula for Pythagorean triplets is correct but also an overkill.

answered Jun 05 '22 at 15:37

dezdichado

13,888

Wow, I totally overkilled it. Thanks! – santm Jun 05 '22 at 15:52

Find all integer solutions, $(ax-b)^2+(bx-a)^2=x$

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