The question is to find all solutions $(x,y)$, where $x$ and $y$ are integers, which satisfy $$ y^k=x^2+x $$ for some integer $k\geq2$. For background, this is from a Putnam type course my university runs, in which we solve problems to prepare for the exam.

Here is what I have. First of all, if $f(x)=x^2+x$, then $f(x)=f(-x-1)$, so we can restrict our attention to $x\geq0$. Now assume $k=2$. In this case we have the easy solutions $(0,0)$ and $(-1,0)$, and I claim there are none others, since $x^2<x^2+x<(x+1)^2$ for $x\geq1$, and hence this will never be a perfect square. Now if $k>2$ is even, there can be no solutions other than $(0,0)$ and $(-1,0)$, since if $(x,y)$ was a different solution, we would have $$ y^{2j}=x^2+x\text{ or }(y^j)^2=x^2+x, $$ implying a different solution to the $k=2$ case.

So we are left with the case where $k$ is odd. If $y$ is odd, then $y^k$ will be odd also, but $x^2+x$ is always even, so there can be no solutions of this form. At this point, I am stuck. I want to say that the only solutions are still $(0,0)$ and $(-1,0)$, but I cannot rule out the case where $y$ is even and $k$ is odd. Could someone help me out here? Also, is there a simpler way to do this that doesn't involve breaking everything down into cases as I have done?

Thanks for your help!


If we factor $x^2 + x = x(x+1)$, we note that the two factors are relatively prime. Now consider $$ y^k = x(x+1), $$ and suppose $p$ is any prime with $p^i$ the highest power of $p$ dividing $y$. Then either $p^{ik}$ divides $x$ with no more factors of $p$, or $p^{ik}$ divides $x+1$ with no more factors of $p$. Therefore, $x$ and $x+1$ are both perfect $k$th powers.

But the only pairs of adjacent perfect $k$th powers, for $k \ge 2$, are $0$ and $1$ and $-1$ and $0$. So $x = -1$ or $0$. In either case, $y = 0$.


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