find the some of two consecutive postive integers sum of whose squares

1.	find the some of two consecutive postive integers sum of whose squares is 365\xa0
Answer» Let one of the consecutive number be x.Therefore other consecutive number be (x + 1)According to question,(x)2\xa0+ (x + 1)2\xa0= 365x2\xa0+ x2\xa0+ 2x + 1 = 3652x2\xa0+ 2x - 364 = 0x2\xa0+ x - 182 = 0x2\xa0+ 14x - 13x - 182 = 0x(x + 14) - 13(x + 14) = 0(x + 14)(x - 13) = 0x = -14 (neglected), x = 13Therefore required two positive integers are 13 and 14.

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