Can someone help me to prove that $\displaystyle n^2+3n+5$ is not divisible by 121 for any $\displaystyle n \in Z$ ?
Suppose that for some $\displaystyle n \in \mathbb{Z}$ that $\displaystyle n^2+3n+5$ is divisible by $\displaystyle 121$Originally Posted by OReilly
then the exists a $\displaystyle \kappa$ such that:
$\displaystyle
n^2+3n+5=121\ \kappa
$
So from the quadratic formula we have:
$\displaystyle
n=\frac{-3 \pm \sqrt{9-4(5-121\ \kappa})}{2}
$
Which implies that $\displaystyle \sqrt{9-4(5-121\ \kappa})$ is an odd integer, $\displaystyle m$ say. So:
$\displaystyle
121\times4\times \kappa - 11=m^2
$
but the LHS is divisible by $\displaystyle 11$, and as $\displaystyle m^2$ is a square it is also divisible
by $\displaystyle 121$ (which is $\displaystyle 11^2$), but that would imply that $\displaystyle 11$ is divisible by $\displaystyle 121$; a
contradiction, so no such $\displaystyle n$ exits.
RonL
If $\displaystyle X = n^2+3n+5$ is divisible by 121 then so is $\displaystyle 4X = 4n^2+12n+20 = (2n+3)^2 + 11$. If $\displaystyle 4X$ is divisible by 121 then it's divisible by 11 and hence 11 divides $\displaystyle (2n+3)^2$. If 11 divides a perfect square $\displaystyle (2n+3)^2$ then it divides $\displaystyle 2n+3$: and that implies that 121 divides $\displaystyle (2n+3)^2$. So we have 121 dividing $\displaystyle (2n+3)^2$ and $\displaystyle (2n+3)^2+11$, hence the difference, which is 11. But that's a contradiction.Can someone help me to prove that $\displaystyle n^2+3n+5$ is not divisible by 121 for any $\displaystyle n \in Z$ ?
If a square is divisible by a prime, it is also divisible by the square of theCan you just explain me this part, I didn't quite understand it.
Originally Posted by CaptainBlack
prime. It's a consequence of the fundamental theorem of arithmetic, that
the decomposition of a number into a product of primes is essential unique.
So the left hand side of the equation being divisible by 11 implies that the
right hand side is also divisible by 11, but the right hand side is a square.
So the right hand side is divisible by 121 (the square of 11). This implies
that the left hand side is divisible by 121, and so that 11 is divisible by
121 which is a contradiction.
RonL
When I was trying to solve the problem I have done this:
$\displaystyle \begin{array}{l}
n^2 + 3n + 5 = 121k \\
n^2 - 8n + 11n + 16 - 11 = 121k \\
(n - 4)^2 + 11n - 11 = 121k \\
(n - 4)^2 = 121k - 11n + 11 \\
(n - 4)^2 = 11(11k - n + 1) \\
\end{array}
$
But I have stoped there.
Let me see if I have understood good.
$\displaystyle 11(11k - n + 1)$ is obviosly divisible by 11, so $\displaystyle (n - 4)^2 $ is also divisible by 11. But $\displaystyle (n - 4)^2 $ is also divisible by $\displaystyle 11^2 $ which now implies that $\displaystyle 11(11k - n + 1)$ is divisible by 121. So we get that $\displaystyle \frac{{11}}{{121}} = \frac{{11k - n + 1}}{{(n - 4)^2 }}$ which shows that 11 is not divisible by 121.
Is that correct?
I didn't know that if a square is divisible by a prime, it is also divisible by the square of the prime. Boy, I feel stupid and embarrassed now!
Divisibility of numbers is simply killing me, I just can't solve any harder problem. I am not expert, but I think that in order to solve those problems you need to know Number theory. Since I am on level of Algebra 1 and 2 for high school (problem that I have posted is from Algebra 1) I find that many harder problems constructed for learning Algebra 1 and 2 that concerns divisibility are based on writers thorought knowledge of Number theory. I don't know if I am wrong, but that's my impression. Either that, or I must find excuse!
Proof,Originally Posted by OReilly
Let $\displaystyle p|ab$ then, $\displaystyle p|a$ or $\displaystyle p\not | a$. If $\displaystyle p\not |a$ then, $\displaystyle \gcd(a,p)=1$ then, $\displaystyle p|b$ by Euclid's Lemma. Thus, $\displaystyle p|a \mbox{ or }p|b$.
Thus, given $\displaystyle p|a^2$ you have, $\displaystyle p|aa$ thus, $\displaystyle p|a$ or $\displaystyle p|a$ thus, $\displaystyle p|a$ definetly. Now, if $\displaystyle c|d$ then, $\displaystyle c^n|d^n$ but since $\displaystyle p|a$ then $\displaystyle p^2|a^2$ if you want to be formal about it
That is right keep on feeling stupid.Originally Posted by OReilly
I don't see how this last equation follows from what you say it does.When I was trying to solve the problem I have done this:
$\displaystyle \begin{array}{l}
n^2 + 3n + 5 = 121k \\
n^2 - 8n + 11n + 16 - 11 = 121k \\
(n - 4)^2 + 11n - 11 = 121k \\
(n - 4)^2 = 121k - 11n + 11 \\
(n - 4)^2 = 11(11k - n + 1) \\
\end{array}
$
But I have stoped there.
Let me see if I have understood good.
$\displaystyle 11(11k - n + 1)$ is obviosly divisible by 11, so $\displaystyle (n - 4)^2 $ is also divisible by 11. But $\displaystyle (n - 4)^2 $ is also divisible by $\displaystyle 11^2 $ which now implies that $\displaystyle 11(11k - n + 1)$ is divisible by 121. So we get that $\displaystyle \frac{{11}}{{121}} = \frac{{11k - n + 1}}{{(n - 4)^2 }}$
RonLwhich shows that 11 is not divisible by 121.
Is that correct?
I didn't know that if a square is divisible by a prime, it is also divisible by the square of the prime. Boy, I feel stupid and embarrassed now!
My lack of understanding again!Originally Posted by CaptainBlack
Let me try again.
From my example we have:
$\displaystyle \begin{array}{l}
(n - 4)^2 - 11 = 121k - 11n \\
(n - 4)^2 - 11 = 11(11k - n) \\
\end{array}
$
That implies that $\displaystyle (n - 4)^2 - 11$ is divisible by 11. But $\displaystyle (n - 4)^2$ is also divisible by 121 but we would have that $\displaystyle (n - 4)^2 - 11$ is also divisble by 121, which is different.
Is that ok?
Not necessarily, you proof falls:
Note if $\displaystyle (n-4)^2=121$ then, $\displaystyle (n-4)^2-11=110$ which is not divisible by 121. Thus, that statement is false.But $\displaystyle (n-4)^2$ is also divisible by 121 but we would have that $\displaystyle (n-4)^2-11$.is also divisble by 121
If,
$\displaystyle a|b,a\not |c$ then, $\displaystyle a\not |(b+c)$
According to you,
$\displaystyle a|b,a\not |c$ then, $\displaystyle a|(b+c)$
Where,
$\displaystyle a=121$
$\displaystyle b=(n-4)^2$
$\displaystyle c=11$
Again in Line,
But $\displaystyle (n-4)^2$ is also divisible by 121 but we would have that $\displaystyle (n-4)^2-11$.is also divisble by 121
From the equation,
$\displaystyle
\begin{array}{l} (n - 4)^2 - 11 = 121k - 11n \\ (n - 4)^2 - 11 = 11(11k - n) \\ \end{array}
$
Your proof.
1)RHS is divisible by 11,
2)So LHS is divisible by 11,
3)$\displaystyle 11|((n-4)^2-11)$
4)Since, $\displaystyle 11|(-11)$ so, $\displaystyle 11|(n-4)^2$
5)So, 121 divides $\displaystyle (n-4)^2$
6)But then, 121 divides 11.
7)Contradiction.
Explain to me step 6!