Divisibility by 121

**ThePerfectHacker** · May 21st 2006, 05:34 PM

Originally Posted by OReilly

121 doesn't divide 11, obviously.
I don't know what I have to explain?

Let me explain to you how a "assume-contradiction" proof work like. If assume something is true, then using a series of conclusion you arrive at a statement which is contradictory. But you need to explain how you arrived at your contradictory statement.

Follow the link of logical statements.
1)We have this statement.
2)We deduce this one.
3)This one is just expressed in mathematical terms.
4)Because if $a|(b+c)$ with $a|b$ then $a|c$ .
5)Because a prime divides a square (explained before).
6)???????????

Note, from steps 1 though 5 each one was a deduction on the one before. IF YOU CAN deduce step six from the previous 5 then you reach step 6 which is a contradiction. Which means your initial hypothesis (your original problem) was false. But to do that you need to demonstrate how step six is related to the other five. You failed to do that. You used wrong theorems on divisibility. However if you CAN provide a logical deduction on step 6 then you arrive at a contradiction and you complete your proof.
--------
Here is an example of what you are doing.
I am going to prove there are infinitely many primes.

1)Assume there are finitely many primes.
2)2=4
3)Contradiction.
4)Thus there must be infinitely many primes.

Note, Step 2 CANNOT BE DEDUCED from step 1 thus it was a faulty prove. You are doing the same, you are faulty believing that you can deduce step 6 from step 5 which is false there is not such theorem on divisibility.
----
In basic terms your proof is wrong.

**OReilly** · May 22nd 2006, 01:13 AM

$ \begin{array}{l} (n - 4)^2 - 11 = 121k - 11n \\ (n - 4)^2 - 11 = 11(11k - n) \\ \end{array} $

We have that $(n - 4)^2$ is divisible by 11. But $(n - 4)^2$ is also divisible by 121 but then we would have that $(n - 4)^2 - 11$ is divisible by 121 which is not true.

Becuase of ${(n - 4)^2 - 11}$ is not divisible by 121 for any $n \in Z$ we have prove that the beginning statement that there is $n \in Z$ for which $n^2 + 3n + 5$ is divisible by 121 is false.

**CaptainBlack** · May 22nd 2006, 02:44 AM

Originally Posted by OReilly

$ \begin{array}{l} (n - 4)^2 - 11 = 121k - 11n \\ (n - 4)^2 - 11 = 11(11k - n) \\ \end{array} $

We have that $(n - 4)^2$ is divisible by 11. But $(n - 4)^2$ is also divisible by 121 but then we would have that $(n - 4)^2 - 11$ is divisible by 121 which is not true.

$(n - 4)^2 - 11$ is divisible by 121 does not follow from: $(n - 4)^2$ is divisible by 121.

For instance suppose $n=15$ then $(n - 4)^2 =121$ and so is divisible by 121, but:

$(n - 4)^2 - 11 =110$ ,

which is divisible by 11 but not by 121.

Becuase of ${(n - 4)^2 - 11}$ is not divisible by 121 for any $n \in Z$ we have prove that the beginning statement that there is $n \in Z$ for which $n^2 + 3n + 5$ is divisible by 121 is false.

**ThePerfectHacker** · May 22nd 2006, 02:16 PM

Is there an echo in here, post #10

**Soroban** · May 28th 2006, 07:34 AM

Hello, OReilly!

Can someone help me to prove that $n^2+3n+5$ is not divisible by 121 for any $n \in Z$ ?

Suppose $n^2+3n+5$ is divisible by 121.

Then: $n^2+3n+5\;=\;121k$ for some integer $k.$

We have the quadratic: . $n^2 + 3n + (5 - 121k)\;=\;0$

Quadratic Formula: . $n \;= \;\frac{-3 \pm \sqrt{3^2 - 4(1)(5 - 121k)}}{2} \;= \;\frac{-3 \pm\sqrt{11(44k - 1)}}{2} $

Since $n$ must be an integer, the discriminant must be a square.
. . That is: $44k - 1$ must be a multiple of $11.$

This is clearly impossible since $\frac{44k - 1}{11}$ will always have a remainder of $10.$

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