Question 1209696: Find all positive integers n for which n^2 - 19n + 59 + n^2 + 4n - 31 is a perfect square.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the given expression be equal to a perfect square $k^2$, where $k$ is a non-negative integer.
We have
\begin{align*} \label{eq:1} n^2 - 19n + 59 + n^2 + 4n - 31 &= k^2 \\ 2n^2 - 15n + 28 &= k^2\end{align*}
Multiplying by 8 on both sides, we have
\begin{align*} 16n^2 - 120n + 224 &= 8k^2 \\ (4n)^2 - 2(4n)(15) + 15^2 - 15^2 + 224 &= 8k^2 \\ (4n - 15)^2 - 225 + 224 &= 8k^2 \\ (4n - 15)^2 - 1 &= 8k^2 \\ (4n - 15)^2 - 8k^2 &= 1\end{align*}
Let $x = 4n - 15$. Then we have the Pell equation $x^2 - 8k^2 = 1$.
The fundamental solution is $(x_1, k_1) = (3, 1)$.
The general solution is given by $x_n + k_n\sqrt{8} = (3 + \sqrt{8})^n$.
When $n = 1$, we have $x_1 = 3$ and $k_1 = 1$.
$4n - 15 = 3 \implies 4n = 18 \implies n = 4.5$, which is not an integer.
When $n = 2$, we have $x_2 + k_2\sqrt{8} = (3 + \sqrt{8})^2 = 9 + 8 + 6\sqrt{8} = 17 + 6\sqrt{8}$.
$x_2 = 17$, $k_2 = 6$.
$4n - 15 = 17 \implies 4n = 32 \implies n = 8$.
$2(8^2) - 15(8) + 28 = 128 - 120 + 28 = 36 = 6^2$.
When $n = 3$, we have $x_3 + k_3\sqrt{8} = (3 + \sqrt{8})^3 = (3 + \sqrt{8})(17 + 6\sqrt{8}) = 51 + 18\sqrt{8} + 17\sqrt{8} + 48 = 99 + 35\sqrt{8}$.
$x_3 = 99$, $k_3 = 35$.
$4n - 15 = 99 \implies 4n = 114 \implies n = 28.5$, not an integer.
When $n = 4$, we have $x_4 + k_4\sqrt{8} = (3 + \sqrt{8})^4 = (17+6\sqrt{8})^2 = 289 + 72 + 204\sqrt{8} = 361 + 204\sqrt{8}$.
$x_4 = 361$, $k_4 = 204$.
$4n - 15 = 361 \implies 4n = 376 \implies n = 94$.
$2(94^2) - 15(94) + 28 = 17672 - 1410 + 28 = 16290$, not a perfect square.
When $n = 3$, $2n^2 - 15n + 28 = 18 - 45 + 28 = 1$.
$n = 3$ is a solution.
When $n = 4$, $2n^2 - 15n + 28 = 32 - 60 + 28 = 0$.
$n = 4$ is a solution.
Final Answer: The final answer is $\boxed{3, 4, 8}$
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