The sum of the squares of two consecutive positive integers is 290. Find the integers.
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Let the two integers be $n$ and $n + 1$. We have that$$n^2 + (n + 1)^2 = 290.$$Simplifying the left side of the equation gives$$n^2 + n^2 + 2n + 1 = 290,$$or$$2n^2 + 2n - 289 = 0.$$Writing this as a quadratic equation gives$$n^2 + n - 144.5 = 0.$$If this has rational roots, they must be of the form $\pm \frac mn$ (where $m$ and $n$ have no common factors), by the Rational Root Theorem. Therefore, the product $mn$ divides $144.5$, which is half of an odd integer. Therefore, we must have that $m = 1$ or $m = 29$, by the prime factorization of $144.5 = 2 \cdot 5 \cdot 29$. In the first case, we get $n^2 + n - 144.5 = 0$, which does not have any integer roots (you can verify this via the quadratic formula), so we can throw it out. In the second case, we get that $$
29n^2 + 29n - 1445 = 0$$and dividing by $29$ gives$$n^2 + n - 50 = 0.$$This quadratic polynomial does have integer roots, namely $n = 5$ and $n = -6$. We must have $n > 0$, so $n = \boxed{5}$.
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