Rijul saini wrote: Indian Postal Training Set 2, Problem 2 Call a triple $(a, b, c)$ of positive integers a nice triple if $a, b, c$ forms a non-decreasing arithmetic progression, $gcd(b, a) = gcd(b, c) = 1$ and the product $abc$ is a perfect square. Prove that given a nice triple, there exists some other nice triple having at least one element common with the given triple. Let us first find all nice triples. A nice triple is $(a-b,a,a+b)$ with $b\ge 0$ and $\gcd(a,b)=1$ and $a(a^2-b^2)$ perfect square. Since $\gcd(a,a^2-b^2)=1$, we get that $a$ and $a^2-b^2$ both are perfect squares. Since $\gcd(a-b,a+b)\in\{1,2\}$, $a^2-b^2$ perfect square implies both $a-b$ and $a+b$ are perfect squares or twice a perfect square. So the nice triples are : either $(m^2,\frac{m^2+n^2}2,n^2)$ with $m^2+n^2=2p^2$ either $(2m^2,m^2+n^2,2n^2)$ with $m^2+n^2=p^2$ Both equations $m^2+n^2=p^2$ and $m^2+n^2=2p^2$ are classical and we get the two solutions : $(2(u^2-v^2)^2,(u^2+v^2)^2,2(2uv)^2)$ (or the reverse) where $u,v>0$ and $u+v$ is odd and $\gcd(u,v)=1$ $((u^2+2uv-v^2)^2,(u^2+v^2)^2,(v^2+2uv-u^2))$ (or the reverse) where $u,v>0$ and $u+v$ is odd and $\gcd(u,v)=1$ plus the previous line with $u=1$ and $v=0$ which gives the solution $(1,1,1)$ Hence the answer : Nice triple $(1,1,1)$ gives other nice triple $(1,25,49)$ for example Nice triple $(2(u^2-v^2)^2,(u^2+v^2)^2,2(2uv)^2)$ (or the reverse) where $u,v>0$ and $u+v$ is odd and $\gcd(u,v)=1$ gives other nice triple $((u^2+2uv-v^2)^2,(u^2+v^2)^2,(v^2+2uv-u^2))$ (or the reverse). Nice triple $((u^2+2uv-v^2)^2,(u^2+v^2)^2,(v^2+2uv-u^2))$ (or the reverse) where $u,v>0$ and $u+v$ is odd and $\gcd(u,v)=1$ gives other nice triple $(2(u^2-v^2)^2,(u^2+v^2)^2,2(2uv)^2)$ (or the reverse).