Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square.

Question

I attempted this by setting $f(n) = \,m^2.\,$

So $\,n^2 + 9n + 30 = m ^2\,$.

Then $\,9(n + 10/3) = (m + n)(m - n)\,$.

So $m = 10/3$ and $n = -17/3$ which is incorrect.

Answer 1

If $n^2+an+b = m^2$ then $4n^2+4an+4b = 4m^2$.

Since $(2n+a)^2 =4n^2+4an+a^2 $, this means that $(2n+a)^2-a^2+4b = 4m^2 $ or $a^2-4b =(2n+a)^2-4m^2 =(2n+a-2m)(2n+a+2m) $.

Looking at all the possible factorizations of $a^2-4b$ (positive and negative), we can get the possible values of $n$ and $m$.

If $a^2-4b = uv$, then $2n+a-2m = u$ and $2n+a+2m = v$ so $4m = v-u$, $4n = v+u-2a$ or $m = \dfrac{v-u}{4}$ and $n = \dfrac{v+u-2a}{4}$.

Since we can consider only non-negative values of $m$, we can restrict the factorization to have $v \ge u$.

Note that this requires both $v-u$ and $v+u-2a$ to be multiples of $4$, which further restricts the solutions.

Also note that since $a^2-4b = uv =(-v)(-u) $, if $m = \dfrac{v-u}{4}$ and $n = \dfrac{v+u-2a}{4}$ is a solution then another solution is $m' = \dfrac{-u-v}{4}=-m$ and $n' = \dfrac{-u-v-2a}{4} = -\dfrac{u+v+2a}{4} = -\dfrac{u+v-2a}{4}-\dfrac{4a}{4} = -n-a $.

In this case, $a=9$ and $b=30$ so $a^2-4b = 81-120 = -39$.

Since $39 = 3\cdot 13$, the possible factorizations (with $v \ge u$) are $(u, v)= (-1, 39), (-3, 13), (-13, 3), (-39, 1)$.

For these $(v-u, v+u-2a)=(v-u, v+u-18)= (40, 20), (16, -8), (16, -28), (40, -56)$.

This are all divisible by $4$, so we get $(m, n)= (10, 5), (4, -2), (4, -7), (10, -14)$.

As expected, if $n$ is a solution, then so is $-n-9$.

Answer 2

Let's assume that $n$ is positive. Observe that $$(n+4)^2=n^2+8n+16<n^2+9n+30<n^2+12n+36=(n+6)^2$$ and the only possibility is $n^2+9n+30=(n+5)^2$.

Now, let's assume that $m=-n$ is positive and $m^2-9m+30$ is perfect square. Also observe that$$(m-5)^2=m^2-10m+25<m^2-9m+30$$

and when $m$ is large enough, $m^2-9m+30<m^2-8m+16=(m-4)^2$ and you can check the case $m^2-9m+30 \ge (m-4)^2$ manually.

Answer 3

If $n>0$ then

By comparison we have $$ (n+4)^2 <n^2+9n+30 < (n+6)^2$$ Therefore $$n^2+9n+30=(n+5)^2$$ Thus $$n=5$$

If $n\leq0$, similarly $$(n+5)^2<n^2+9n+30< (n+4)^2$$ for all $n<-14$

Now we only need to consider the case $-14\leq n \leq 0$

Manually, it leads to $n=-14$, $n=-7$ and $n=-2$

Answer 4

Hint:

Let $n^2+9n+30=(n+a)^2$ where $a$ is any integer

$\iff n=\dfrac{a^2-30}{9-2a}$ which has to be an integer

Now if integer $d(>0)$ divides both $a^2-30,9-2a;$

$d$ must divide $2(a^2-30)+a(9-2a)=9a-60$

$d$ must divide $-9(9-2a)-2(9a-60)=39$

So, a necessary condition for $(9-2a)|(a^2-30)$ is $9-2a$ must divide $39$