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I just finished the following question:

Find all positive integers $N$ such that $N\times 2^{N+1}+1$ is a perfect square

I solved it in the following way:

$N\times2^{N+1}+1=k^2$ for $k\in Z$

$N\times2^{N+1}=(k-1)(k+1)$

So we have that $(k-1)(k+1)\equiv0 \pmod 2$, so $k+1 \equiv0\pmod2$ and $k-1\equiv0\pmod2$.

Since $k+1$ and $k-1$ differ by just 2, and they are both even, then one of the two is a multiple of $2^b$, where $b\in Z$ and $b>1$ and the other if divided by two, the result is odd. From this and the fact that $k+1>k-1$ we have that:

$k+1=2^N$ and $k-1=2N$ (since $2^N>2N$).

So $2N+2=2^N$ for which $N=3$ is the only solution. This can be proved as follows:

We have $f$ such that $f(x)=2x+2$, so $f'(x)=2$ and $g$ such that $g(x)=2^n$, so $g'(x)=\ln(2)\times2^x>2$

So the only solution is for $N=3$.

I find my logic overly complex and laborious, could you please suggest some alternative approaches?

Adam Zalcman
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  • By the way I believe that my solution is correct, however if you find some error, please tell me about it –  Sep 08 '20 at 08:25
  • Is it necessary that $(k+1)(k-1)\equiv 0$ mod $2\Rightarrow $ $(k+1)\equiv 0$ mod $2$ and $(k-1)\equiv 0$ mod $2$? – Kumar Sep 08 '20 at 08:30
  • Yes because that means that either $k+1\equiv 0mod2$ or $k-1\equiv 0mod2$, and if one is even then the other is even, as there difference is 2, which is an even number –  Sep 08 '20 at 08:32
  • Immediately reduces to the linked problem (see answer below). – Bill Dubuque Oct 30 '23 at 14:11

1 Answers1

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Other apporoach can be $N \times 2^{N+1} + 1 = k^2$. Hence,$k$ is in the form of $2m+1$ and we will have: $N \times 2^{N+1} + 1 = 4m^2 +‌4m + 1$. Then, $N \times 2^{N+1} = 4 \times m \times (m+1) \Rightarrow N \times 2^{N-1} = m \times (m+1)$. Now, we have a simple proof that $gcd(m, m+1) = 1$, and one of them is odd and te other is even. Now we have the following case:

$N$ is odd

  • ($m = N$) $N = 2^{N-1} - 1 \Rightarrow N+1 = 2^{N-1}$. As $\lim_{N\to\infty}\frac{N+1}{2^{N-1}} = 0 $, and for all $N >‌ 3$, $N + 1 < 2^{N-1}$ (by induction), the only answer is $N = 3$.
  • ($m = 2^{N-1}$) $2^{N-1} = N-1$. The same as the foregoing case, for all $N > 3$, $N - 1 < 2^{N-1}$. Hence, we can cannot find any solution here.

$N$ is even

We can factorize $N = 2^{\alpha_0} q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_i^{\alpha_i}$. However, like the previous cases, if we decompose the left side to two even and odd numbers, we will reach an inequality such that one side of it is greater than $2^{N-1}$, and the other side is less than $N-1$. Therefore, in that case, we will find $N \leq 3$ as well. The only even number in this range is 2, but it does not satisfy the condition. Hence, there is no answer when $N$ is even.

OmG
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  • For the case where $N$ is odd, you can conclude that $2^{N+1}$ is a square as well. Since $N \cdot 2^{N+1}$ is a perfect square, $N \cdot 2^{N+1}+1$ can't be a perfect square. – Haran Sep 08 '20 at 10:20