(Q6)Find two consecutive positive integers, sum of whose squares is 613.

Answer

Let a positive integer be x.

Then the second integer = x + 1

Sum of the squares of the above integers = x2 + (x+1)2

= x2 + x2 + 2x + 1

= 2x2 + 2x + 1

By problem 2x2 + 2x + 1 = 613

2x2 + 2x - 612 = 0

x2 + x - 306 = 0

x(x + 18) - (x + 18) = 0

(x - )(x + 18) = 0

x - = 0 (or) x + 18 = 0

x = or -18,

we do not consider -18

Then the numbers are (, + 1)

, are the required two consecutive positive integers.