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Find the value(s) of positive integer $n$ such that $n² + 19n + 48$ is a perfect square.

I factorised it to $(n+3)(n+16)$, but that gives negative integer answers $-3$ and $-16$. What do I do?

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Notice that since:

$$n^2+19n+48=(n+3)(n+16)\Rightarrow $$$$(n+3)^2<(n+3)(n+16)<(n+16)^2 $$

This means that you must check only a finite number of cases:

$$n^2+19n+48=(n+k)^2 \ \ \ k\in\{4,5,...,14,15\}$$ $$19n+48=2kn+k^2$$ $$n=\frac{k^2-48}{19-2k}$$

Since $n$ is positive clearly $k\leq 9$ and notice that we have:

$$19-2k\leq k^2-48 \Rightarrow k\geq 8$$

So we must check only $k=8$ and $k=9$:

$$k=8 \Rightarrow n=\frac{16}{3}$$ $$k=9 \Rightarrow n=33$$

So the only solution is $n=33$

:)

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  • $\begingroup$ Thanks bro 😁😁😁 $\endgroup$ – user664431 May 2 at 8:05
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$n^2+19n+48=m^2$, $4n^2+76n+192=4m^2$, $(2n+19)^2-361+192=4m^2$, $(2n+19)^2-4m^2=169$, $(2n+19+2m)(2n+19-2m)=169=13^2$, so for each way of factoring $169$ you get a system of two linear equations in $m$ and $n$. Can you take it from there?

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  • $\begingroup$ Yeah sure... Thanks for your help 😁😁😁😁😁😁👍👍👍👍 $\endgroup$ – user664431 May 1 at 9:29
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Hint:

For $n\in[1,5]$, $(n+7)^2<n^2+19n+48<(n+8)^2$.

For $n\in[6,33)$, $(n+8)^2<n^2+19n+48<(n+9)^2$.

For $n\in(33,\infty)$, $(n+9)^2<n^2+19n+48<(n+10)^2$.

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