Finding the smallest $n$ such that $n^{2}$ ends with $00001$ The problem is to find the smallest natural number $n$ so that its square's last $5$ digits are $00001$. $n$ and $n^{2}$ cannot begin with $0$. I know that we have a lower bound on $n$, namely $\sqrt{100001}≈316.23$. Intuitively, I think the smallest natural number with the property has to be $100001$ and I have not found any counterexamples to this (in fact I checked using some code and it seems to be correct). I haven't been successful in proving this rigorously however. Do I use the lower bound in some way? How to approach this type of questions in general, i.e. how do digits respond to squaring/cubing? All help is appreciated.
Find all solutions to $x^2\equiv 1\pmod {91},\ 91 = 7\cdot 13$ I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.  For $x^2\equiv 1\pmod {7}$, i did: $$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.  For $x^2\equiv 1\pmod {13}$, i did: $$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.  Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
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