# Is $n^2+3n+6$ divisible by 25, where $n$ is a integer?

If we want $$n^2+3n+6$$ to be divisible by $$25$$, it firstly has to be divisible by $$5$$. So, let's take a look at couple of cases: $$n=5k: 25k^2+15k+6$$. The remainder is $$6$$, so it's not divisible by $$25$$.

$$n=5k+1: 25k^2+25k+10$$. The remainder is $$10$$, so it is not divisible by $$25$$.

And so on for $$n=5k+2, n=5k+3, n=5k+4.$$

1. Is this a good way to do this?

The other way that came to my mind would be following:

Let's say that $$n^2+3n+6$$ is divisible by $$25$$: $$n^2+3n+6=25k$$ $$n^2+3n+(6-25k)=0$$ If we solve this equation, we get $$n_{1,2}=\frac{-3 \pm \sqrt{5} \sqrt{20k-3}}{2}.$$ But $$\sqrt{5}$$ is a irrational number, so is $$n$$ irrational number too. This is contradiction, so it's not divisible by $$25$$.

1. Is this a good way to solve this problem?
• Your second method needs to consider $\sqrt{20k-3}$ too as that might balance the $\sqrt{5}$ and make the overall result rational - in fact it does not, but you have not shown it does not – Henry Mar 11 at 9:58

## 2 Answers

If $$25$$ divides $$n^2+3n+6,$$ the later will be divisible by $$5$$

$$n^2+3n+6\equiv n^2-2n+1\pmod5$$

So, we need $$n\equiv1\pmod5\implies n=1+5m$$

$$(5m+1)^2+3(5m+1)+6=25m^2+25m+10\not\equiv0\pmod{25}$$

• @Dietrich, extra dot – lab bhattacharjee Mar 11 at 13:13

Hint: We have $$n^2+3n+6=(n+4)^2 \bmod 5$$

Now look at Bill's answer at this duplicate:

$n^2 + 3n +5$ is not divisible by $121$