Find two consecutive positive odd integers whose square has the sum of 290?

Answer 1

Let n be the first odd positive integer
Let n + 2 be the second odd positive integer

n² + (n+2)² = 290
n² + (n² + 4n + 4) = 290
2n² + 4n + 4 = 290
n² + 2n + 2 = 145
n² + 2n - 143 = 0
(n + 13)(n - 11) = 0

So n = 11 or -13, but only the first answer satisfies the given condition of being positive.

The consecutive odd integers are: 11 and 13.

(Double-check: 11² + 13² = 121 + 169 = 290)

Note: you can also get to this intuitively without doing all the algebra. What's half of 290? It is 145. This is the average of the smaller and larger square. The closest square to 145 is 12². So intuitively you should realize that the odd integers will be 11 and 13.

Answer 2

11 & 13

11^2 + 13^2 = 290