Does an integer $9
Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.
This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks.

We are solving $n(n-1)=n^2-n\equiv0\pmod{100}$. Since $\gcd(n,n-1)=1$, one of $n$ or $n-1$ must be a multiple of $4$ while the other must be a multiple of $25$.This leads to the equations
$$
\begin{align}
4x-25y=+1\tag{1}\\
4x-25y=-1\tag{2}
\end{align}
$$
For $(1)$, $n=4x$ and $n-1=25y$. For $(2)$, $n=25y$ and $n-1=4x$.
Using the Euclidean algorithm, $(1)$ has solutions $(x,y)=(-6+25k,-1+4k)$ and $(2)$ has solutions $(6+25k,1+4k)$. The two solutions that give $4x$ and $25y$ between $9$ and $99$ are $(19,3)$ and $(6,1)$.
$(19,3)$ solves $(1)$ so $n=4x=76$ and $76^2=5776\equiv76\pmod{100}$
$(6,1)$ solves $(2)$ so $n=25y=25$ and $25^2=625\equiv25\pmod{100}$
Thus, the two integers that satisfy the given condition are $25$ and $76$.
