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

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
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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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