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?

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 $$n=33$$

:)

• Thanks bro 😁😁😁 – user664431 May 2 at 8:05

$$n^2+19n+48=m^2$$, $$4n^2+76n+192=4m^2$$, $$(2n+19)^2-361+192=4m^2$$, $$(2n+19)^2-4m^2=169$$, $$(2n+19+2m)(2n+19-2m)=169=13^2$$, 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?

• Yeah sure... Thanks for your help 😁😁😁😁😁😁👍👍👍👍 – user664431 May 1 at 9:29

Hint:

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

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

For $$n\in(33,\infty)$$, $$(n+9)^2.