The sum of the squares of two consecutive multiples of 7 is 637 . Find the multiples .

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Solution

Let the multiples of 7x&7(x+1)
$(7 x)^{2} + (7 (x + 1))^{2} = 637$
$\Rightarrow 49 x^{2} + 49 (x + 1)^{2} = 637$
$\Rightarrow x^{2} + (x + 1)^{2} = 13$
$\Rightarrow x^{2} + x^{2} + 1 + 2 x - 13 = 0$
$\Rightarrow 2 x^{2} + 2 x - 12 = 0$
$\Rightarrow x^{2} + x - 6 = 0$
$\Rightarrow (x + 3) (x - 2) = 0$
so x=-3 or x=2
no s= -14,-21 no s=14,21

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