Here goes (this may not be a very direct proof, and I haven't looked at the hints):
The square of an odd number can be written as follows:
(2n+1)^2
Since any odd number must be one more than a previous even number, and the previous even number must be divisible by 2.
This can be expanded out and factorised to:
4(n^2+n)+1
n^2+n will always be an even number, since if n is even, n^2 will be even, and n will be even (even+even=even). If n is odd, n^2 will be odd, and n will be odd (odd+odd=even).
Since n^2+n is always even, it is a multiple of 2. 4*(Some Multiple of 2) is a multiple of 8.
ie. 4(n^2+n) will always be a multiple of 8. Since there is a +1 term on the end of the expression, the result will always be one more than a multiple of 8.
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