# Find two consecutive positive integers, the sum of whose squares is 365.

Let the first number be ‘X’, so the other number will be ’(X+1)’.

X2+(X+1)2 = 365

X2 + X2 + 1 + 2X – 365 = 0

2X2 + 2X – 364 = 0

X2 + X – 182 = 0

On applying Sreedhracharya formula

X = –14 or X = 13.

The numbers are 13 & 14.

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