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# An Olympiad Problem

Show that the equation has infinite integer solution : $$x^2 + y^2 + z^2 = (x - y)(y - z)(z - x)$$.

Note by Dev Sharma
2 years, 7 months ago

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I started the question by applying the substitution $$y=x+a, z=x+a+b$$. We will restrict $$a,b$$ to be positive integers. This will give us a quadratic in $$x$$ which I omit here. My first thought, a lucky one, was to look at the case $$a=b$$: $$3x^2+6ax+5a^2-2a^3$$. After the quadratic we have:

$$x=-a\pm \frac{a\sqrt {6(a-1)}}{3}$$.

We want $$x$$ to be an integer, so $$a-1=6n^2$$ for some integer $$n$$.

Hence $$x=(6n^2+1)(-1\pm 2n)$$, $$y=(6n^2+1)(-1\pm 2n)+6n^2+1, z=(6n^2+1)(-1\pm 2n)+2(6n^2+1)$$. $$n$$ is an integer so we can generate an infinite number of integer solutions.

In all the proving infinite solution problems I've encountered in the past, a majority of them required looking at special cases; this should be a natural approach especially for an olympiad problem. The most complicated one I've dealt with involved special cases and pell's equation, can't remember the exact problem tho.

- 2 years, 7 months ago

Thanks.

but why did you make y = x + a ....? and why only one case a = b??

- 2 years, 7 months ago

the substitution is motivated by the product of differences on the right. I only examine the $$a=b$$ because that's all it suffices to show the infinitude of the solutions.

- 2 years, 7 months ago

Interesting question. What have you tried?

Staff - 2 years, 7 months ago

first of all, i want to thank you as you edited my que.. I tried but couldn't get any pattern.. And nobody is giving solution.

- 2 years, 7 months ago

Try listing out a few small solutions. E.g. $$(0, 0, 0), (-1, 0, 1), ( -1, 1, 2), (-2, -1, 1 )$$. Are there any others?

Staff - 2 years, 7 months ago