Prove (or disprove): For positive integer $n$, define $M_n$ as an $n$-digit positive integer such that in decimal representation, the last $n$ (rightmost) digits of $M_n^2$ is $M_n$ itself. Denote $L_n$ as the sum of all possible values of $M_n$. Then the longest string of consecutive integers $k$ that satisfy the equation $L_k=10^k+1$ is $6$.
As an explicit example: If $n=3$, then $M_3 = 376,625$ because $376^2=141\underline{376}$ and $625^2=390\underline{625}$ and then $L_3=10^3+1 $ but if $n=4$, $M_4=9376$ only, so $L_4\ne10^4+1$.
My attempt: By Chinese Remainder Theorem, I'm able to obtain that the first few values of $k$ the satisfy the condition $L_k =10^k + 1$ are $3,6,7,8,9,10,11,14,15,16,17,18,19,21,22,23,26,27,28,30,31,36,37,38,39,41$.
I'm curious though: why does so many values of natural number $k$ satisfy the equation $L_k=10^k+1$? I estimate it's about $66\frac23\%$ of all natural numbers.
If $M_n^2\equiv M_n\pmod{10^n}$, then with $N_n:=10^n+1-M_n$ we have $N_n^2\equiv (1-M_n)^2\equiv 1-2M_n+M_n^2\equiv 1-M_n\equiv N_n\pmod{10^n}$. Therefore such numbers always come in pairs that add up to $10^n+1$. There are always exactly two $M_n$ because we can solve exactly for the next digit to obtain a $M_{n+1}$ from an $M_n$: With the ansatz $M_{n+1}=M_n+d(10^n)$ we find $M_{n+1}^2=M_n^2+2dM_n10^n+d^210^{2n}$ and this is $\equiv M_{n+1}\pmod{10^{n+1}}$ iff $(2M_n-1)d\bmod {10}$ is a specific value. And as $2M_n-1\equiv \pm1\pmod{10}$, there exists exactly one digit $d$ that fits. This way we obtain exactly two infinite digit sequences such that any "tail" of them is a $M_n$: $$\ldots6109004106619977392256259918212890625 $$ and $$\ldots 3890995893380022607743740081787109376$$
Your point that $L_4\ne 10^4+1$ comes from the fact that you do not allow leading zeroes; otherwise $0625^2=39\underline{0625}$ leads to $L_4=10^4+1$ as well. Thus we have $L_k\ne 10^k+1$ iff at least one of the digit sequences has a $0$ in the $k$th place from the right, or equivalently: the first sequence has a $0$ or a $9$ in the $k$th place. If we assume for the moment that the digits are random, then this leads to a probability of $80\,\%$ instead of $66\frac23\,\%$. It is not an exact prove of such random-compatible behaviour of the numbers in question, but the statistics for the first 10000 digits produces 967 times the digit $0$ and 963 times digit $9$ (for the sequence ending in $625$), hence $L_k=10^k+1$ holds $80.7\,\%$ of the time for this range.
As a sidenote, there is a string of length $33$ (so much more than $6$) of consecutive integers $k$ with $L_k=10^k+1$ (somewhere around $k=8800$); and if we assume random nature of the digits again, arbitrarily long such strings can be expected.