<s>[INST] Fake induction proofs Question: Can you provide an example of a claim where the base case holds but there is a subtle flaw in the inductive step that leads to a fake proof of a clearly erroneous result? [Note: Please do not answer with the very common all horses are the same color example.]
Comment: Sometimes inductive arguments can lead to controversial conclusions, such as the surprise exam paradox, Richard's paradox and a host of other paradoxes. However, I am interested in examples of a more mathematical nature (as opposed to linguistic) where the inductive argument is subtly flawed and leads to erroneous conclusions. 
Note: If you provide an answer, please do so in a way similar to how current answers are displayed (gray out the flaw so people can be challenged to discover it).
 [/INST] Claim:
If $a$ is an odd square modulo $m$, then $a$ is a square modulo $2m$.
Proof:
Let $m=2^v k$ where $k$ is $m$'s largest odd factor. Proceed by induction on $v$.
Base case:
$v=0$, so $m=k$ is odd.
$a$ is a square modulo $k$, so, for some $x$, $k\mid x^2-a$. Modulo $2k$,
\begin{align*}
(k+x)^2 &= k^2+2kx+x^2\\
&= k+x^2\tag{$k^2=k$ as $k$ is odd; $2kx=0$}
\end{align*}
so $x^2$ and $(k+x)^2$ are $a$ and $k+a$ in some order. So $a$ is a square modulo $2k$.
Inductive step:
$a$ is a square modulo $2^v k$ so there is an $x$ where $x^2=a\mod 2^v k$.
As the base case is proved, we may suppose that $m$ is even and $v\geqslant1$.
This means that, as $a$ is odd, $x$ is odd.
Knowing that $m$ is even also enables us to refine $x$:
Modulo $m$,
\begin{align*}
(m/2+x)^2 &= (2^{v-1} k+x)^2\\
&= 2^{2v-2} k^2+2\cdot2^{v-1} kx+x^2\\
&=2^{v} kx+x^2\\
&=x^2\\
&=a
\end{align*}
as the other terms are multiples of $m=2^v k$. That is, we may take $0<x<m/2$.
Modulo $2m$,
\begin{align*}
(m/2+x)^2 &= (2^{v-1} k+x)^2\\
&= 2^{2v-2}k^2+2\cdot2^{v-1} kx+x^2\\
&=2^v kx+x^2\\
&=mx+x^2\\
&=m+x^2\tag{as $x$ is odd}
\end{align*}
Thus $x^2$ and $(m/2+x)^2$ are $a$ and $m+a$ in some order. Thus $a$ is a square modulo $2m$. QED.
Flaw:

 The base case is valid. The inductive step relies on $2^{v+1}k\mid2^{2v-2}k^2$, which is true only if $2v-2\geqslant v+1$ i.e. $v\geqslant3$ i.e. $8\mid m$.

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