# There exists $n$ such that there exists an n-digit number M for which $M(10^n + 1)$ is a perfect square.

Show that there exists a positive integer $n$ greater than 3 such that there exists an $n$-digit positive integer $M$ for which $M(10^n + 1)$ is a perfect square.

• My guess was to pick $M=10^n + 1$, for $(n+1)$-digit $M$. Commented Jan 8, 2017 at 10:26
• @user398623 -- What makes you so sure that $10^n + 1$ is prime only when $n = 1$ or $n = 2$? Commented Jan 8, 2017 at 10:33
• For this to work, any prime $p$ such that $p|10^n+1$ and $p^2\not |10^n+1$ must divide $M$. In particular if $10^n+1$ is squarefree, then $M(10^n+1)$ can only be a perfect square if $10^n+1|M$, in which case $M$ is not an $n$-digit number. $10^n+1$ is square-free for $n=3,4,5,6,7,8,9$. Commented Jan 8, 2017 at 10:37
• @ArnaudD. I've edited my question. Commented Jan 8, 2017 at 13:18

As I explained in the comments, this is impossible if $10^n+1$ is squarefree. On the other hand, if $p^2$ divides $10^n+1$ for some prime $p$, then $M_0=\frac{10^n+1} {p^2}$ is an integer, and $M_0(10^n+1)=\left(\frac{10^n+1} {p}\right)^2$ is a perfect square. $M_0$ does not necessarily have $n$ digits; but we can always multiply it by some power of $4$ or $9$ in such a way that the product $M$ has $n$ digits, and then $M$ will satisfy the conditions.
So all we have to do is prove that there is some $n$ for which $10^n+1$ is divisible by the square of a prime number $p$; $p$ cannot be $2$, $3$ or $5$, so we will try $p=7$. We want to have \begin{align}10^n \equiv -1 \pmod{49}.\end{align} By Euler's theorem $$10^{42}\equiv 1 \pmod{49},$$ and thus $$(10^{21}+1)(10^{21}-1)\equiv 0 \pmod{49}.$$ To conclude that $49|10^{21}+1$ it suffices to prove that it does not divide $10^{21}-1$. We have \begin{align}10^{21}\equiv (10^{3})^7\equiv 10^3\equiv 3^3\equiv 27\equiv -1\pmod 7, \end{align}and thus $7\not |10^{21}-1$, which implies that $49\not |10^{21}-1$.