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Prove that $1+8k$, for all nonnegative integers $k$, is a quadratic residue modulo $2^n$, where $n$ is a positive integer.

For $n = 5$, we have $0,1,9,17,25$, which are all quadratic residues modulo $2^5$. How do we prove this in general?

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    If you have seen the result that the multiplicative group $\Bbb{Z}_{2^n}^*$ is isomorphic to $C_2\times C_{2^{n-2}}$, then you know that exactly one quarter of the odd residue classes modulo $2^n$ will be quadratic residues. Obviously the square of an odd number is $\equiv1\pmod8$, so we know which quarter it is! Of course, the fact you want to prove is more or less equivalent to that result about the group structure. – 2017-01-09

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Prove by induction. Suppose that the claim is true for a given $n$. Let $k$ be a nonnegative integer and we are looking for a solution $1+8k\equiv a^2$ mod $2^{n+1}$. Such a solution is also a solution mod $2^n$ which we know that we have by induction, namely there is some $a$ such that $1+8k=a^2+m 2^n$ for some integer $m$. If $m$ is even, then this is also a solution mod $2^{n+1}$ and you are done. Otherwise, note that $$(a+2^{n-1})^2=a^2+a2^n+2^{2n-2}=1+8k+(a-m)2^n+2^{2n-2}$$ Since clearly $a$ must be odd, you get that $(a-m)$ is even, and if you also assume that $n\geq 3$, then $2n-2\geq n+1$ so modulo $2^{n+1}$ you get $1+8k$ which is what you wanted.

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We can assume $n\ge 4$, because otherwise it is trivial.

Fact 1: Two odd numbers have the same square modulo $2^n$ if and only if they are equal or opposite modulo $2^{n-1}$.

Proof: If $(2s+1)^2\equiv (2t+1)^2\pmod {2^n}$ then $s^2+s\equiv t^2+t\pmod {2^{n-2}}$ and $$(s-t)(s+t+1)\equiv 0\pmod{2^{n-2}}$$

Only one of these factors is even, so we have $s\equiv t$ or $s+t\equiv -1 \pmod{2^{n-2}}$ and the fact follows.

An immediate consequence is that each odd square modulo $2^n$ has exactly four square roots.

Then there are exactly $2^{n-3}$ odd quadratic residues modulo $2^n$, a fourth of the odd residues.

On the other hand,

Fact 2: If $r$ is odd and $m\not\equiv 1\pmod 8$ then $r$ is not a quadratic residue modulo $2^n$.

Proof: Suppose the contrary. If $r\equiv m^2\pmod{2^n}$, we have that $2^n$ divides $(m^2-1)-(r-1)$. Since $8$ divides $m^2-1$, it also divides $r-1$, a contradiction.

To finish the proof, note that the set of odd numbers from $0$ to $2^n$ can be divided in $2^{n-2}$ packs of $4$ numbers each. These packs can be of the form $\{8k+1,8k+3,8k+5, 8k+7\}$. Among these four numbers only $8k+1$ can be a square. Since a fourth of the odd residues are quadratic, $8k+1$ must be a square for each $k$.