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The following is the problem and the part of my answer.

$n\in N$, $~n\times 2^{n+1}+1~$ is perfect square.

Find the value of $~n$.

$n\times 2^{n+1}+1=x^{2} \Longrightarrow n\times 2^{n+1}=(x+1)(x-1)$

$x+1=2^{\alpha}a~$, $~x-1=2^{\beta}b~~$

Here I can't go anymore. Thanks in advance.

Bill Dubuque
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chloe_shi
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  • For $n=0$ and $n=3$ we have a perfect square. No idea how we can prove that there are no more. – Peter Oct 30 '23 at 08:58
  • Your thoughts are correct. If we look at $n\cdot 2^{n+1} = (x+1)(x-1)$ we see that both sides are even. And since $x+1$ and $x-1$ are 2 distinct, one of them is divisible by $2$ exactly once and the one is divisible by $2^n$. Suppose that $x + 1 = k\cdot 2^n$ fpr some $k$ (the other case is analogous) – Lereu Oct 30 '23 at 08:59
  • I would post this as an answer, but the powers that be think that if there is one way to approach a problem, no other approaches can possibly be interesting or valid. $n\equiv 1,2 \bmod 3 \Rightarrow n\cdot 2^{n+1}+1 \equiv 2 \bmod 3$, hence not a square. So $n=3k$. This leads to $6k\cdot 8^k =(x-1)(x+1)$. Easy to show that $k=1$ (i.e. $n=3$) is the only positive integer solution. BTW, this approach does not work with the post that is at the root of the "dup" chain used to close this post, so it's hard to imagine that it is a dup. – Keith Backman Oct 31 '23 at 16:08

2 Answers2

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Start by checking $n=0,1,2,3$ which yields two solutions: $(0,1)$ and $(3,7)$.

Going forward consider only $n\geq 4$.

$n \cdot 2^{n+1} = (x-1)(x+1)$

Note that since $2 | (x-1)(x+1)$, both $x-1$ and $x+1$ are even.

$gcd(x-1,x+1) = gcd(x-1,x+1-(x-1)) = gcd(x-1,2) = 2$.

Since $2^{n+1}|(x-1)(x+1) \Rightarrow 2^n|(x-1)$ or $2^n|(x+1)$.

$Case $ $ 2^n|(x-1):$

$2^n|(x-1) \Rightarrow \exists k \in \mathbb{N},$ $ x-1 = k\cdot2^n \iff x = k \cdot 2^n+1.$

$n \cdot 2^{n+1} = (x-1)(x+1) \iff n \cdot 2^{n+1} = k\cdot2^n\cdot(k\cdot2^n+2)\iff n=k\cdot(k\cdot2^{n-1}+1)$.

$n = k\cdot(k\cdot2^{n-1}+1) \geq 2^{n-1}+1 > n$ for $n\geq4$.

$Case $ $ 2^n|(x+1):$

$2^n|(x+1) \Rightarrow \exists k \in \mathbb{N},$ $ x+1 = k\cdot2^n \iff x = k \cdot 2^n-1.$

$n \cdot 2^{n+1} = (x-1)(x+1) \iff n \cdot 2^{n+1} = k\cdot2^n\cdot(k\cdot2^n-2)\iff n=k\cdot(k\cdot2^{n-1}-1)$.

$n = k\cdot(k\cdot2^{n-1}-1) \geq 2^{n-1}-1 > n$ for $n\geq4$.

Hence $(0,1)$ and $(3,7)$ are the only solutions.

Duffoure
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You are on the right track, just take $n=2^pm$ for $p\in\mathbb N$ and $m\in2\mathbb N+1$, then

$$x+1=2^\alpha a\ \ \text{and}\ \ x-1=2^\beta b$$

for some $a,b\in2\mathbb N+1$ with $ab=m$ and $\alpha+\beta=2^pab+1$. Since $2^\alpha a-2^\beta b=2$, we have $\alpha=1$ or $\beta=1$.

  • If $\alpha=1$ then $\beta=2^pab$ and it follows $$2^{2^pab-1}=\frac{a-1}b.$$ But this is impossible since $2^{2^pab-1}\geq2^{a-1}>a-1\geq\frac{a-1}b$ for all $a\geq1$.

  • If $\beta=1$ then $\alpha=2^pab$ and it follows $$2^{2^pab-1}=\frac{b+1}a.$$ We must have $b\leq 3$ because $2^{2^pab-1}\geq2^{b-1}>1+b\geq\frac{b+1}a$ for all $b>3$. This gives $b=1$ or $b=3$, and in both cases we must have $a=1$ (otherwise the RHS is not integer). Hence the only possibilities are $(a,b,p)=(1,1,1)$ or $(1,3,0)$, and you can easily check only the latter is a valid solution, that is, when $n=2^pab=3$.

This shows $n=3$ is the only positive solution, along with $n=0$ you get all the solutions.

Mengchun Zhang
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