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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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    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.2010-11-08
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    @Mike Spivey: Thank you for the warning!2010-11-10
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    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.2010-11-11
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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?$2011-06-18
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    You got a great answer by Bill, when a number is divisible by $121$ what is it congruent to mod $11^2$?2012-05-06

3 Answers 3

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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?2012-05-06
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    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.2012-05-07
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    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* $k$?2012-05-07
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    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}$2012-05-07
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    @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.2012-05-07
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    @AntonioVargas : Ok, let us consider $\sqrt{(44\cdot k)-1}$ is irrational number. Value of $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ will rational or integer only when we will get another $11$ in $\sqrt{(44\cdot k)-1}$, so that even pair of $\sqrt {11}$ will make a $11$. Here $44\cdot k$ is multiple of 11 but $(44\cdot k)-1$ isn't. So we'll not get another $\sqrt {11}$, hence $\sqrt {11}$ will make make whole expression a irrational number.2012-05-07
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    @AntonioVargas : Hope, your all doubts are cleared now.2012-05-07
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    I don't believe the statement "Value of $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ will be rational or integer only when we get another 11 in $\sqrt{(44\cdot k)-1}$." Why can't $\sqrt{(44\cdot k)-1}$ have the form $q/\sqrt{11}$? I don't think your argument is complete, iahmed. Even if you can fill in every gap I point out, how can you prove that there are no more gaps?2012-05-07
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    What you are trying to say is $\sqrt {(44 \cdot k)-1} = q/\sqrt{11}$ where $q$ is rational number. This implies that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1} = q$ but I have already told you that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ is irrational, so how can you get $q$ as rational?2012-05-07
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    iahmed, you have not proved that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ is irrational for every $k$. You have only shown that it is irrational when $\sqrt{(44\cdot k)-1}$ is an integer, and you have shown that it is impossible to have $44k-1 = 11a$, when $a$ is an integer. (Note that you claimed you have shown the case for rational $a$ too, "will rational or integer only when..." but you have not proved it.) You have only shown that it is irrational in some special cases, but you have not shown that those are ALL of the possible cases, (cont.)2012-05-07
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    (cont.) and because of that you cannot say "but I have already told you that $\sqrt{11}\cdot \sqrt{(44\cdot k)-1}$ is irrational". Do you understand?2012-05-07
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    Okay, I'm sorry for that & for using `rational` term everywhere. I just want to prove $(\sqrt{11}\cdot \sqrt{(44\cdot k)-1})$ isn't an integer, may be value of this expression is rational but it can't be integer and if $(\sqrt{11}\cdot \sqrt{(44\cdot k)-1})$ isn't integer, $n$ isn't too. I'm very sorry for considering integer and rational both at a time.2012-05-07
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    That's fine iahmed, but you still have not proved it.2012-05-07
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    I can prove it. If _x_ and _y_ are not perfect squares, for $\sqrt{x}\sqrt{y}$ to be an integer there must exist some integer _n_ such that $x = \frac{n}{2^b}$ and $y = 2^b{n}$, where _b_ is an integer. In this case, that means there must be some multiple of 11 that is equal to half of $\sqrt{44k - 1}$. However, any number that is a multiple of 44 is also a multiple of 11, and subtracting 1 from it means it cannot possibly be a multiple of 11. Therefore there is no number that satisfies that condition. QED2013-05-07
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    Wow, I was really sloppy with that proof. _x_ and _y_ could also be equal, _b_ must be even, and it should be just $44k - 1$ without the radical.2013-05-07
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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 $11^2\not|33$