Do there exist positive integers $n,k$ such that $n^2+4n=k^2$? I'm not sure how to attack this question. I was able to get that $n=\frac{-4\pm\sqrt{16+4k^2}}{2}$, but I don't think it is of any use.
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$\begingroup$Clearly the number is greater then $n^2$, so the smallest square it could be is $(n+1)^2 = n^2 + 2n + 1 \neq n^2 + 4n$ because $2n +1\neq 4n $ for $n \in \mathbb N$.
The next square is $(n+2)^2 = n^2 + 4n + 4$ which is bigger than our number.
So in conclusion no, $n^2 + 4n$ is never a perfect square
$\endgroup$ $\begingroup$HINT: Recall that
$$(n+2)^2=n^2+4n+4$$
$\endgroup$ 1 $\begingroup$We have
$$ n^2 < n^2+4n < (n+2)^2 $$
So if $n^2+4n$ is a square, we must have that $n^2+4n=(n+1)^2$. But this quickly leads to the equation $2n=1$, which is impossible.
$\endgroup$ $\begingroup$Note first that there are no integers between any other two integers.
Now note that $n^2+4n+4 = (n+2)^2$ meaning $n^2+4n = (n+2)^2-4$.
For large enough $n$ this can be bounded between two consecutive squares so you just have to prove these bounds and demonstrate for small $n$ the specified relation holds.
$\endgroup$ $\begingroup$Assume $n^2+4n = k^2 \Rightarrow n(n+4) = k^2$. There are $2$ cases:
Case 1: $ n = m^2, n+4=p^2$ whereas $mp = k \Rightarrow 4 = p^2-m^2 = (p-m)(p+m)$, and you can take it from here
Case 2: $n = a, n+4 = ap^2$ whereas $ap = k \Rightarrow 4 = a(p^2-1) = a(p-1)(p+1)$, and you can also take it from here.
$\endgroup$ 0 $\begingroup$For a number to be a square for any prime divisor $p$ of that number, $p^2$ should also divide it.
Now write $n^2+4n$ as $n(n+4)$; except $2$ no other prime number can divide both $n$ and $n+4$. So if $n$ is not a square, $n^2+4$ cannot be. But if $n$ is a square $n+4$ should also be a square for $n^2+4n$ to be a square. Among positive integers there are exist no two square differing by 4.
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