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October 5, 2016 at 6:13 pm
#337

admin

Keymaster

Actually, Yash your answer is partially correct because it is true that the solutions to $x^2 \equiv 1 \bmod 13$ are only $x \equiv 1,-1 \bmod 13$. However, your proof is incorrect and this may not generalize to all moduli other than $13$. For example, notice that $x^2 \equiv 1 \bmod 15$ also has solutions $x \equiv 4,-4,1,-1 \bmod 15$ so there is some issue with your work. Nonetheless, great start and you have been awarded 15 points, 5 points for this problem and 10 points for the other one.