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.

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.

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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.

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Thanks.

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

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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.

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Interesting question. What have you tried?

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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.

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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?

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