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Question:

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

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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 3

up vote 9 down vote accepted

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

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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 $121$.

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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 $11^2\not|33$

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