Find three consecutive integers such that the sum of their squares is 50.


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Question Number 217326 by Rasheed.Sindhi last updated on 10/Mar/25

$F i n d t h r e e c o n s e c u t i v e i n t e g e r s$ $s u c h t h a t t h e s u m o f t h e i r s q u a r e s$ $i s 50 .$
	Answered by Hanuda354 last updated on 10/Mar/25

	$Let n, n + 1 and n + 2 are required numbers .$ $n^{2} + {(n + 1)}^{2} + {(n + 2)}^{2} = 50$ $3 n^{2} + 6 n + 5 = 50$ $3 n^{2} + 6 n - 45 = 0$ $3 (n^{2} + 2 n - 15) = 0$ $3 (n + 5) (n - 3) = 0$ $n = - 5 \lor n = 3$ $n = - 5 \Rightarrow - 5, - 4, - 3$ $n = 3 \Rightarrow 3, 4, 5$
	Commented by Rasheed.Sindhi last updated on 10/Mar/25

	$T h a n X s i r!$
	Answered by Rasheed.Sindhi last updated on 10/Mar/25

	$b - a = 1 \Rightarrow a^{2} + b^{2} - 2 a b = 1$ $c - b = 1 \Rightarrow b^{2} + c^{2} - 2 b c = 1$ $c - a = 2 \Rightarrow c^{2} + a^{2} - 2 c a = 4$ $Adding :$ $2 ((a^{2} + b^{2} + c^{2}) - (a b + b c + c a)) = 6$ $50 - (a b + b c + c a) = 3$ $a b + b c + c a = 47$ ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$ ${(3 b)}^{2} = 50 + 2 (47)$ $9 b^{2} = 50 + 94 = 144$ $b^{2} = 16$ $b = 4, - 4$ $b - 1 = 3, - 5$ $b + 1 = 5, - 3$
	Answered by mr W last updated on 10/Mar/25

	${(n - 1)}^{2} + n^{2} + {(n + 1)}^{2} = 50$ $n^{2} = 16$ $n = \pm 4$ $\Rightarrow (- 5, - 4, - 3), (3, 4, 5)$
	Commented by Rasheed.Sindhi last updated on 10/Mar/25

	$T han k s sir!$
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