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

Question:

Show that $n^2 + 3n + 5$ is not divisible by $121$, where $n$ is an integer.

• I see you have asked ten questions on this site thus far with having accepted answers to any of them. It is considered polite here to formally accept the best answer you receive for a given question. In case you don't know how to do that: Each answer should have a little check mark by it. Click on the check mark next to the answer you want to accept. – Mike Spivey Nov 8 '10 at 23:59
• @Mike Spivey: Thank you for the warning! – Paulo Argolo Nov 10 '10 at 21:55
• I wouldn't call it a warning. :) Every social group has its own etiquette and social norms; newcomers just have to learn what those are. – Mike Spivey Nov 11 '10 at 0:47
• Isn't the easiest way to show this just to pick, say, $n=0...?$ Or did you mean $n$ such that $n^2 + 3n + 5 > 121?$ – barf Jun 18 '11 at 9:34
• You got a great answer by Bill, when a number is divisible by $121$ what is it congruent to mod $11^2$? – Kirthi Raman May 6 '12 at 20:12

## 3 Answers

HINT $\rm\quad\ m\ =\ n^2 + 3\:n+5\ \equiv\ (n-4)^2\ \:(mod\ 11)\ \Rightarrow\ n\ =\ 4+11\:k \ \Rightarrow\ m = \ldots\ (mod\ 11^2)$

Make a contradiction that $n^2 + 3n + 5$ is divisible by $121$
Let $k$ be any positive integer, we can say that
$n^2 + 3n + 5 = 121\cdot k$
$n^2 + 3n + (5 - (121\cdot k)) = 0$

Solve for $n$,

\begin{align} n=&\frac{-3 \pm \sqrt {(3)^2 - 4\cdot1\cdot(5-(121\cdot k))}}{2\cdot1}\\ n=&\frac{-3 \pm \sqrt {(484\cdot k)-11}}{2} \end{align}

Given that $n$ is an integer, so $\sqrt {(484\cdot k)-11}$ should be an integer
We can represent $\sqrt {(484\cdot k)-11}$ as $(\sqrt{11}\cdot \sqrt{(44\cdot k)-1})$, whose value can't be an integer as value of $\sqrt{11}$ is irrational.
So we can say that our assumption is wrong, $n^2 + 3n + 5$ is not divisible by $121$.

• Why does $\sqrt{11}$ being irrational imply that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ isn't an integer? – Antonio Vargas May 6 '12 at 23:19
• Suppose we choose value of $k$ in such a way that $\sqrt {(44*k)-1}$ is integer. Whenever we will multiply a integer or rational number with irrational, we will always get a result as irrational or non-integer number. – rekenerd May 7 '12 at 7:16
• So what about other values of $k$? How do you know that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ is not an integer for any $k$? – Antonio Vargas May 7 '12 at 7:25
• You take any value of $k$, may be it is possible that we will get integer value of $\sqrt{(44\cdot k)-1}$, but $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ will never be integer as $\sqrt{11}$ is going multiply with $\sqrt{(44\cdot k)-1}$ – rekenerd May 7 '12 at 7:35
• @AntonioVargas : Suppose value of $\sqrt{(44\cdot k)-1}$ is any integer, let say $8$. Vaule of $\sqrt {11}$ is $3.31662479$. When we will do multiplication, we get $26.532998323$, which is irrational number. – rekenerd May 7 '12 at 7:40

As $121=11^2,$ we need $11|(n^2+3n+5)$

Let us find $x,y$ such that $x-y=3,x+y=11\implies x=7,y=4$

$$n^2+3n+5=(n+7)(n-4)+33$$

As $33$ is divisible by $11,$ so must be $(n+7)(n-4)$ to make $11|(n^2+3n+5)$

Now $11|(n-4)\iff 11|(n+7)$ as $(n+7)-(n-4)=11$

So in that case, $11^2|(n+7)(n-4),$ but $11^2\not|33$