Question

15.
Determine the set of integers n for which n2 + 19n +92 is a square.
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Answer 1

Question

$n^2+19n+92$ is a perfect square for some natural numbers. Find all of them.

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Solution

Let's consider $n^2+19n+92=k^2$.

For each factor to be integers, it is required that it is factorized over integers. However, this polynomial for $n$ cannot be. To resolve this we multiply $4$ on both sides.

Then $4n^2+76n+368=4k^2$.

$\implies (2n+19)^2+7=4k^2$

$\implies (2k)^2-(2n+19)^2=7$

$\implies (2k+2n+19)(2k-2n-19)=7$

Each factor is an integer. Now we note that the factors of LHS are two factors of $7$. The followings are four ways to factorize.

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Case 1

$\begin{cases} & 2k+2n+19=7 \\ & 2k-2n-19=1 \end{cases}\implies k=2,n=-8$

Case 2

$\begin{cases} & 2k+2n+19=-7 \\ & 2k-2n-19=-1 \end{cases}\implies k=-2,n=-11$

Case 3

$\begin{cases} & 2k+2n+19=1 \\ & 2k-2n-19=7 \end{cases}\implies k=2,n=-11$

Case 4

$\begin{cases} & 2k+2n+19=-1 \\ & 2k-2n-19=-7 \end{cases}\implies k=-2,n=-8$

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The set of integers $n$ is $\{-11,-8\}$. This is the required answer.

Answer 2

$\huge\bf\fbox\red{Answer:-}$

Let's consider $n^2+19n+92=k^2$

For each factor to be integers, it is required that it is factorized over integers. However, this polynomial for nn cannot be. To resolve this we multiply 44 on both sides.

Then $4n^2+76n+368=4k^2$

$\implies (2n+19)^2+7=4k^2$

$\implies (2k)^2-(2n+19)^2=7$

$\implies (2k+2n+19)(2k-2n-19)=7$

Each factor is an integer. Now we note that the factors of LHS are two factors of 7 . The followings are four ways to factorize.

Case 1

$\begin{gathered}\begin{cases} & 2k+2n+19=7 \\ & 2k-2n-19=1 \end{cases}\implies k=2,n=-8\end{gathered}$

Case 2

$\begin{gathered}\begin{cases} & 2k+2n+19=-7 \\ & 2k-2n-19=-1 \end{cases}\implies k=-2,n=-11\end{gathered}$

Case 3

$\begin{gathered}\begin{cases} & 2k+2n+19=1 \\ & 2k-2n-19=7 \end{cases}\implies k=2,n=-11\end{gathered}$

Case 4

$\begin{gathered}\begin{cases} & 2k+2n+19=-1 \\ & 2k-2n-19=-7 \end{cases}\implies k=-2,n=-8\end{gathered}$

The set of integers n is $\{-11,-8\}$.

This is the required answer.