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.

(In reply to

Puzzle solution by K Sengupta)

SOLUTION TO PART B:

When n = p(p+1)/2, then n is called a Triangular Number.

Putting p = 1,2,3,.... in turn, the first few triangular numbers (excluding zero) are :

1, 3, 6, 10, 15, 21, .....

Since, the Triangular numbers are of the form p(p+1)/2, the ith Triangular Number will always be equal to the sum of the first i natural numbers.

Accordingly:

1 = 1

3 = 1+2

6 = 1+2+3

10 = 1+2+3+4

15 = 1+2+3+4+5