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.
odd number is of the form 2*k+1
its square=4*(kČ+k)+1=4*k*(k+1)+1
k*(k+1) is definitely even bcoz either k or (k+1) is divisible by 2.
Let k*(k+1)=2*n for some n,
then square=8*n+1. Hence Proved
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Posted by Praneeth
on 2007-08-01 13:11:24 |
Solution
