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Question

Question 4
Find two consecutive positive integers, sum of whose squares is 365.

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Solution

Let the consecutive positive integers be x and x+1.
Therefore, x2+(x+1)2=365
x2+x2+1+2x=365
2x2+2x364=0
x2+x182=0
(Splitting the middle term x as 14x-13x because 14x×13x=182x2)
x2+14x13x182=0
x(x+14)13(x+14)=0
(x+14)(x13)=0
Either, x+14=0orx13=0,
x=14 or x=13
Since the integers are positive, x can only be 13.
x+1=13+1=14
Therefore, two consecutive positive integers will be 13 and 14.

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