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

(Q) Find the two consecutive positive integers, sum of whose squares is 365.

Ans: Let’s the 1st number be $x$

Next number has to be consecutive and hence next number will be $x + 1$

It is given that sum of the squares of these two number is 365

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

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

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

$\Rightarrow 2 x^2 + 2x - 364 = 0$

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

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

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

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

$\Rightarrow x = -14 ~ and ~ x = 13$

Here, we reject x = – 14 because it is given that the numbers need to be positive.

Therefore, $x = 13, x + 1 = 13 + 1 = 14$

Hence, the two numbers are 13 and 14.

Check:

If numbers are 13 and 14 then let’s check the given condition:

$(13)^2 + (14)^2 = 169 + 196 = 365$

Since it meets the given condition, hence our answer is correct.

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