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What are all integer $(n,m)$, such that $n^3+2^m\cdot n$ is a perfect square

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Insights, background, self effort, ideas...?? – DonAntonio Aug 22 '12 at 3:59
Welcome to MSE: since you are new, I wanted to let you know a few things about the site. First, you might want to read the faq. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – mixedmath Aug 22 '12 at 4:01

I’ve left a bit for you to finish, but here’s a pretty detailed guide to a solution.

Notice that $n^3+2^mn=n(n^2+2^m)$, which is a perfect square iff either (1) $n$ and $n^2+2^m$ are both perfect squares, or (2) $n^2+2^m=na^2$ for some positive integer $a$.

In case (1) say $n=a^2$ and $n^2+2^m=b^2$. Then $2^m=b^2-a^4=(b-a^2)(b+a^2)$, and $b-a^2$ and $b+a^2$ must both be powers of $2$; say $b-a^2=2^r$ and $b+a^2=2^s$, where $r+s=m$ and $r<s$. Then $a^2=2^{s-1}-2^{r-1}=2^{r-1}(2^{s-r}-1)$, which is a perfect square iff $r-1$ is even and $2^{s-r}-1$ is a perfect square (why?). Let $t=s-r\ge 1$; for what positive integers $t$ is $2^t-1$ a perfect square? If $t>1$, then $4\mid 2^t$, so $2^t-1\equiv 3\pmod 4$. But $2^t-1$ is odd, and every odd square is congruent to $1$ mod $4$, so we must have $t=1$. Thus, $r$ must be odd, $s=r+1$, and $m=2r+1$.

Conversely, let $r=2k+1$ be any odd positive integer, and let $m=2r+1$. Let $n=2^{r-1}=2^{2k}$. Then $n^3+2^mn=2^{6k}+2^{m+2k}=2^{6k}+2^{6k+3}=2^{6k}(1+2^3)=9\cdot2^{6k}=(3\cdot2^{3k})^2$. This gives us one infinite family of solutions.

In case (2) we must have $n\mid 2^m$, in which case $n=2^k$ for some $k\in\{0,1,\dots,m\}$. Then $na^2=2^{2k}+2^m=2^k(2^k+2^{m-k})$, so $a^2=2^k+2^{m-k}$. If $m=2k$, $a^2=2\cdot2^k=2^{k+1}$, so $k$ must be odd. Conversely, each odd value of $k$ yields a solution of this type; what are these solutions?

If $m\ne 2k$, $a^2=2^k+2^{m-k}$ is the sum of two different powers of $2$, so we need to know when such a sum is a perfect square. Consider the sum $2^r+2^s$, where $r$ and $s$ are non-negative integers, and $r<s$. If $r$ is even, then $2^r+2^s=2^r(1+2^{s-r})$ is a perfect square iff $1+2^{s-r}$ is. If $r$ is odd, then $2^r+2^s=2\cdot2^{r-1}(1+2^{s-r})$ is a perfect square iff $2(1+2^{s-r})$ is. But $2(1+2^{s-r})$ cannot be a perfect square (why?), so we need only consider the case of even $r$ and ask when $1+2^t$ is a perfect square, where $t=s-r\ge 1$. Certainly this is the case when $t=3$. Are there any other solutions?

Suppose that $1+2^t$ is a perfect square, say $1+2^t=c^2$. Then $2^t=c^2-1=(c+1)(c-1)$. What does this tell you about the numbers $c+1$ and $c-1$?

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