Find two consecutive odd positive integers, sum of whose squares
is 290
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11 and 13
ANSWER
Let x an odd positive integer
Then, according to question
x2+(x+2) 2 =290
2x2+4x−286=0
x2+2x−143=0
x2 +13x−11x−143=0
(x+13)(x−11)=0
x=11 as x is positive
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