Very easy to see (and well known) that if a\equiv s^2 (mod 8) than for each k tjere exist r such that a\equiv r^2 (mod 2^k) (If a is a quadratic res mod 8 than it is a quadratic res mod 2^k) -7 -- quadratic mod 8.

Cool! Here's what I did: I used induction. Let f(n) be the greatest power of 2 which divides n. We prove the property for k=1 and then we assume it's true for k from 1 to t and we try to prove it for t+1: There are infinitely many b s.t. b²+7=n*2^t for some n. Let's assume the sentence is false for t+1, so there are infinitely many b s.t. f(b²+7)=2^t. We can obviously find infinitely many a s.t. f(a(a+2b))=2^t (we just take a s.t. f(a)=2^t-1 ). We get that f(b²+a(a+2b))=2^t+1, so f((a+b)²)=2^t+1, so we have a contradiction because we assumed the sentence was false for k=t+1, so the sentence is true for k=t+1, and the induction is over.

A silly overkill with Hensel's Lemma.It suffices to show there is at least one integer $x$ for each $k$ such that $n\cdot 2^k-7=x^2(\star)$.See that if $x$ works then $(2^k+x)$ works too.Since $(2^k+x)^2=2^k(2^k+2x)+x^2=(n+2^k+2x)\cdot 2^k-7$ We consider the polynomial $f(x)=x^2+7$. We will try to find a root of $f(x)$ modulo $2$.To rephrase: $x^2+7\equiv 0\pmod{2}$ we get $x=1$ works.But here $f^{\prime}(x)\equiv 2x \equiv 0\pmod{2}$ so Hensel's Lemma doesnot directly work.Indeed $f(x)=x^2+7\equiv (x-1)^2$ in $\mathbb{F}_2[x]$.But see that $f(1)=8=4\cdot 2$ so for each $k>1$ there is a lifting of the root modulo $2^k$.Hence for each $k$ there exists an integer $x_k$ such that $x_k^2+7\equiv 0\pmod{2^k}\implies x_k^2+7=n_k\cdot 2^k$ and hence we are done.Because we wanted to find (at least) one integer $x$ such that $(\star)$ is satisfied.

Set $a=7a', b=7b'$ which gives $7a'-b'2^k=-1$ Thus if we can show that perfect squares can be written in the form $b'2^k$ the result immediately follows by Pell But this is trivial to show so we're done

Math1331Math wrote: Set $a=7a', b=7b'$ which gives $7a'-b'2^k=-1$ Thus if we can show that perfect squares can be written in the form $b'2^k$ the result immediately follows by Pell But this is trivial to show so we're done I too got sorta same.. but is it correct?

it may be but it is incomplete. this only tests for multiples of $7$.

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All numbers of the form $((2^k-7)^2)m^2$are solutions .Or I am not understanding the question.

The problem is equivalent to showing that for a fixed $k,$ there are infinitely many positive integers $a,$ such that $2^k|a^2+7$. It is obvious for $k=1, 2, 3$ that we can just choose $a \equiv 1 (\mod 8)$. Assume now the statement is true for all integers up to some $k$. We shall prove it for $k+1$. Let's look at some $a$ that is solution of the divisibility for $k$. Obviously $a$ is odd. If $2^{k+1}|a^2+7$, we can just pick the same $a$ as a solution for $k+1$. Let's see what happens if $2^k$ divides $a^2+7,$ but $2^{k+1}$ doesn't. We want to find an integer $r,$ such that $a+r$ is a solution. In other words, we want $a^2+7+2ar+r^2$ to be divisible by $2^{k+1}$. Note that $r$ must be even. We have $r(r+2a) = r^2+2ar \equiv-a^2-7 \equiv 2^k (\mod 2^{k+1})$. Since $a$ is odd, $r+2a$ will be only divisible by $2$ and so we should look for $r$ divisible by $2^{k-1},$ but not by $2^k$. So we decide to go with $r=2^{k-1}$. Now $a+r$ is indeed a solution, since $(a+r)^2+7 = a^2+7+r^2+2ar \equiv 2^k+0+2^k \equiv 0 (\mod 2^{k+1})$ and we are done $\blacksquare$. Remark: For the $r^2$ term to vanish, we need $2k-2 \geq k+1,$ or $k \geq 3$, so for our solution it is crucial to check the cases $k=1, 2, 3$. One can easily prove wrong statements like: For all $k,$ there is an integer $a,$ such that $a^2+3$ is divisible by $2^k$ with the same lifting method if he or she doesn't check the base cases.

Claim: Suppose $x^2\equiv -7\pmod{2^k}$ for some $k\ge 3$. Then \[ x^2\equiv -7 \pmod{2^{k+1}} \quad \text{or} \quad (x+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}.\]Proof: We know $\ell:= \tfrac{x^2+7}{2^k} \in \mathbb{Z}$ from the condition. Case 1: $\ell$ is even. Then \[\frac{\ell}{2} = \frac{x^2+7}{2^{k+1}} \in \mathbb{Z},\]so $x^2\equiv -7 \pmod{2^{k+1}}$. Case 2: $\ell$ is odd. Then \[ \frac{x^2+2^kx+7}{2^{k+1}} = \frac{x^2+7}{2^{k+1}} + \frac{x}{2} =\frac{\ell+x}{2} \in \mathbb{Z}\]since $x$ and $\ell$ are both odd. Therefore, $-7\equiv x^2+2^kx \equiv (x+2^{k-1})^2\pmod{2^{k+1}}$. This proves the claim. $\blacksquare$ We have $(1+8\ell)^2\equiv -7\pmod8$ for any $\ell\ge 0$. So we can build up an infinite family of pairwise distinct values using the claim that are $-7\pmod{2^k}$, increasing the exponent of the power of $2$ one each time, by initially having the $\ell$'s sufficiently far apart.

Remark

We show by induction on $k$ that there exists a positive integer $z$ for which $z^2 \equiv -7 \pmod{2^{k}}$. The problem statement will follow then as $z+r\cdot 2^{k}$ will also work for any $r \in \mathbb{Z^{+}}$ Base Case:- for $k=1,2,3$ we can take $z=1$ and base case trivially works. Now, Claim:- suppose $z^2 \equiv -7 \pmod{2^{k}}$ for $k \geqslant 3$ . Then: $ z^2\equiv -7 \pmod{2^{k+1}} \quad \text{or} \quad (z+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}.$ Proof:- set $\mathcal{F}=\frac{z^2+7}{2^{k}} \in \mathbb{Z}$ Case 1:- if $\mathcal{F}$ is even then, $\frac{\mathcal{F}}{2} \in \mathbb{Z} \implies \frac{z^2+7}{2^{k+1}} \in \mathbb{Z}$ which gives $z^2 \equiv -7 \pmod{2^{k+1}}$ Case 2:- if $\mathcal{F}$ is odd. Then we notice: $\frac{z^2+7}{2^{k+1}}+\frac{z}{2}=\frac{\mathcal{F}+z}{2} \in \mathbb{Z}$ as $\mathcal{F}$ and $z$ are odd. hence we get $\frac{z^2+2^{k}z+7}{2^{k+1}} \in \mathbb{Z}$ i.e. $z^2+2^{k}z \equiv -7 \pmod{2^{k+1}} \implies z^2+2^{k}z+2^{k+1}\cdot 2^{k-3} \equiv -7 \pmod{2^{k+1}} \implies (z+2^{k-1})^2 \equiv -7 \pmod{2^{k+1}}$ hence claim proved $\blacksquare$ now clearly from our setup of $z+r\cdot 2^{k}$ we get infinite such square numbers as it's true for any $r \in \mathbb{Z^{+}}$ $\blacksquare$

We wanna prove that $a^2\equiv -7\pmod{2^k}$ has solutions for all positive integers $k$. For $k=1$, $a\equiv 1\pmod{2}$ For $k=2$, $a\equiv 1,3\pmod{4}$ For $k=3$, $a\equiv 1,3,5,7\pmod{8}$ For $k=4$, $a\equiv 3,5,11,13\pmod{16}$ For $k=5$, $a\equiv 5,11,21,27\pmod{32}$ For $k=6$, $a\equiv 11,21,43,53\pmod{64}$ For $k=7$, $a\equiv 11,53,75,117\pmod{128}$ Notice that $(2^{k-1}-a)^2\equiv a^2\pmod{2^k}$. Also notice that $a$ must always be odd. If $a^2\equiv -7\pmod{2^{k+1}}$, then we're done by induction. Otherwise, $(2^{k-1}-a)^2\equiv a^2-2^ka\equiv -7\pmod{2^{k+1}}$, and we're done by induction.