Take any odd number and square it. It will invariably be a multiple of 8 plus 1. So (odd)^2=8n+1 where n is an integer. Show why this is always so. Also show what the pattern for n is.
Any odd number can be expressed as 2n+1
So, (odd)^2 = (2n+1)^2 = 4(n^2+n) +1 = 4n(n+1) +1
Now, n(n+1) is always even, irrespective of whether n is even or odd.
Therefore, m(m+1) =2*n, where m is a integer.
Thus, (odd)^2 = 8*n+1, where m is an integer.
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