# MOG 2017 4

We are given that n is odd \therefore let n=2N+1 where N is an integer greater than 1 (N\in\mathbb{Z} | N>1).
\therefore M=(2N+1)^{2} + 2(2N+1)-7
= 4N^2+4N+1+4N+2-7
=4N^2 + 8N-4
=2(2N^2+4N-2)=4(N^2+2N-1)
It looks like weâ€™ve found our 4 factors of M: 2,4,(2N^2+4N-2),(N^2+2N-1)
But what if two of these factors are congruent or equal to 1 for some N? This would mean that we had in fact only found 2 or 3 factors of M and not completed the proof. To complete the proof we need to show that none of 1,2,4, (2N^2+4N-2),(N^2+2N-1) are equal for any N.

Case 1: 2=2N^2 +4N-2
N^2 +2N-2=0
N=-1\pm\sqrt{3} which is not an integer \therefore we never have Case 1

Case 2: 2=N^2+2N-1
N^2+2N-3=0
N=1 and N=-3. This isnâ€™t a problem though because N>1, so Case 2 doesnâ€™t happen

Case 3: 4=2N^2 +4N-2
N^2+2N-3=0
This is the same as Case 2

Case 4: 4=N^2 +2N-1
N^2+2N-5=0
N=-1\pm\sqrt{6}

Case 5: 2N^2 +4N-2=N^2+2N-1
N^2+2N-1\neq0 for an integral N so we can divide by this
2=1 clearly not true so finally Case 5 doesnâ€™t happen

Case 6 2N^2+4N-2=1
2N^2+4N-2=1
2N^2+4N-3=0
N=\frac{-2\pm\sqrt{10}}{2} not an integerâ€¦

Case 7 N^2+2N-1=1
N^2+2N-1=1
N^2+2N-2=0
This is the same as Case 1

Only now can we say that we have found 4 distinct factors of M as we have shown that none of them are equal for integral N.

We donâ€™t have to find the factors in terms of n, we just have to show that 4 distinct factors exist which we have done. So if you want, you can find our factors in terms of n but it isnâ€™t necessary.

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