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**Logic** This happens when p(p-1) is zero, i.e. p = 0 or p = 1. 0 is not a natural number so p = 1 is a solution and c is, indeed, 4.

However, I am unsure how to prove that $\displaystyle p(p-1)=k^2$ can never occur for any integer k bigger or equal to 0.

I have come up with this:

Let$\displaystyle p(p-1)=k^2$, then $\displaystyle p^2-p-k^2=0$. The discriminant is $\displaystyle D=1+4k^2$... which should not be an exact square? It can never equal 4, 9, 16, etc., I mean. Is this so?