Find $n^2+58n$, such that it is a square number

Question

I have the following problem. Find all numbers natural n, such that $n^2+58n$ is a square number.

My first idea was, $n^2+58n=m^2$

$58n=(m-n)(m+n)$ such that $m-n$ or $m+n$ must be divisible by 29, but this leads to nothing.

Answer 1

Another hint. You can look at the problem as a quadratic equation $$x^2+58x-m^2=0$$ with $\Delta=58^2+4m^2$ and (excluding negative $n$'s) $$n=\frac{-58+2\sqrt{29^2+m^2}}{2}=-29+\sqrt{29^2+m^2}$$ which reduces to finding integers of the forms $29^2+m^2=q^2$, which is easier since $29$ is a prime and $$29^2=(q-m)(q+m)$$ allows for the following cases

• $q-m=29$ and $q+m=29$
• $q-m=1$ and $q+m=29^2$
• $q-m=29^2$ and $q+m=1$

Answer 2

Hint: Write your equation in the form $$(n+29)^2=x^2+29^2$$

Answer 3

Hint: \begin{eqnarray}k^2 &=& n^2+58n \\ &=& \underbrace{n^2+2\cdot 29n+\color{red}{29^2}}\color{red}{-29^2} \\ &=& (n+29)^2-29^2\end{eqnarray}

so $$ 29^2 = (n+29)^2-k^2 $$ $$29^2= (n+29+k)(n+29-k)$$

Answer 4

$(n+29)^2=x^2+29^2$ so there is a Pythagorean triple $(A,29,B)$ so you have $$29=m^2-n^2\Rightarrow (m,n)=(15,14)$$ Then $$n+29=15^2+14^2\Rightarrow n=392$$ Thus $$392^2+58\cdot392=176400=420^2$$ You can verify that the solution is unique: $n=392$