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

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

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TopNewestI 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. – Xuming Liang · 2 years, 1 month ago

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but why did you make y = x + a ....? and why only one case a = b?? – Dev Sharma · 2 years, 1 month ago

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– Xuming Liang · 2 years, 1 month 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.Log in to reply

Interesting question. What have you tried? – Calvin Lin Staff · 2 years, 1 month ago

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– Dev Sharma · 2 years, 1 month 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.Log in to reply

– Calvin Lin Staff · 2 years, 1 month 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?Log in to reply