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

Q4.    Find two consecutive positive integers, the sum of whose squares is 365.

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Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of squares is 365.

.$x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

$\Rightarrow x^2 - 13x+14x+182 = 0$

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$.

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