# MOG 2015 5

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@mon1 got there just before me as I was typing this up but I thought I might as well post it

Part (a)
Let x^2=N and x\equiv r \pmod8 so that x=8q+r where 0\leqslant r < 8\!.

\begin{align} N&=x^2\\ &=(8q+r)^2\\ &=64q^2+16rq+r^2\\\\ \therefore N&\equiv r^2 \pmod{16} \end{align}

Checking the remainders of 0^2, 1^2,\cdots,7^2 upon division by 16\!, we see that the only options are 0, 1, 4 and 9\!.

Part (b)
Taking the equation \!\!\pmod{16} and using the result from part (a):

\begin{align} m!+12&\equiv 0,1,4 \text{ or } 9\\ m!&\equiv 4,5,9 \text{ or } 13\\ \end{align}

Note that m! for m\geqslant6 is divisible by 6!=720=16\times45\!, and is therefore equivalent to 0\pmod{16} and does not work.

Testing m from 1 to 5, we find that the only solutions are (m, n)=(4, 10),(5, 14)\!.

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@mon1 @arctangent MULTIPLE SOLUTIONS ARE THE DREAM

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