Can someone show me the proof by mathematical induction that $\displaystyle n^2+n+2$ isn't divisible by 49 for every n (natural number).

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- Aug 23rd 2006, 06:45 PMOReillyDivisible by 49
Can someone show me the proof by mathematical induction that $\displaystyle n^2+n+2$ isn't divisible by 49 for every n (natural number).

- Aug 23rd 2006, 06:55 PMThePerfectHackerQuote:

Originally Posted by**OReilly**

- Aug 23rd 2006, 07:09 PMOReillyQuote:

Originally Posted by**ThePerfectHacker**

- Aug 23rd 2006, 07:23 PMThePerfectHackerQuote:

Originally Posted by**OReilly**

I was just writing out a proof. But I accidently closed my internet thing and lost everything I typed. - Aug 23rd 2006, 10:32 PMrgep
Firstly observe that n^2+n+2 is divisible by 7 if and only if n is congruent to 3 modulo 7: that is, if n is of the form 7x+3. See this by writing it as n^2 - 6n + 9 + 7(n-1), so that it is (n-3)^2 modulo 7.

Now substitute n=7x+3 into n^2 + n + 2 to get 49x^2 + 49x + 14, which cannot be divisible by 49. - Aug 24th 2006, 06:23 AMOReilly
Is this proof valid?

Supose that for $\displaystyle n=k$ expression is divisible by 49:

$\displaystyle k(k + 1) + 2 = 49a$

Supose that now for $\displaystyle n=k+1$ expression is also divisible by 49:

$\displaystyle (k + 1)(k + 2) + 2 = 49b$

From first equation we get that $\displaystyle k + 1 = \frac{{49a - 2}}{k}$.

Substituting into second equation we get:

$\displaystyle \begin{array}{l}

(\frac{{49a - 2}}{k})(k + 2) + 2 = 49b \\

\frac{{49ak + 98a - 2k - 4}}{k} = 49b - 2 \\

49ak + 98a - 2k - 4 = 49bk - 2k \\

49ak + 98a - 2k - 4 - 49bk + 2k = 0 \\

49ak + 98a - 49bk - 4 = 0 \\

\end{array}

$

Last expression isn't divisible by 49 so $\displaystyle n^2+n+2$ isn't divisible by 49. - Aug 24th 2006, 02:40 PMThePerfectHacker
You need to show that,

If,

$\displaystyle n^2+n+2\not = k$

then,

$\displaystyle n^2+n+2\not = k+1$

Equivalently (contropositive),

If,

$\displaystyle n^2+n+2=k+1$

then,

$\displaystyle n^2+n+2=k$

You shown that:

if,

$\displaystyle n^2+n+2=k$

then,

$\displaystyle n^2+n+2\not = k+1$

This is not equivalent to the above two statements. - Aug 24th 2006, 05:10 PMQuickI don't get it...
I don't understand your proof.

where am I off?

If $\displaystyle n^2+n+2=k$

then, $\displaystyle n^2+n+3=k+1$

:confused: - Aug 24th 2006, 05:57 PMThePerfectHacker
Here is how you do it.

---

You need to show there doth not exist such a $\displaystyle 49k$ such as,

$\displaystyle n^2+n+(2-49k)=0$

Which means there are no integral solution for $\displaystyle n$.

If you solve for "n",

$\displaystyle n=\frac{-1\pm \sqrt{1-4(2-49k)}}{2}$.

Note the following, if the expression in the radical is a square then "n" is an integral number because the square root of an odd square is an odd whole number and when added of subtracted to 1 produces an even number and when finally divided by two produces an integral number, thus, if

$\displaystyle 1-4(2-49k)=196k-7$ is not a square then,

$\displaystyle 1-4(2-49(k+1))=1-4(-47-49k)=196k+189$ is also not a square.

Thus, again you need to show that.

IF,

$\displaystyle 196k-7$ is not a square

THEN,

$\displaystyle 196k+188$ is not a square.

~~Contropositive Time~~

IF,

$\displaystyle 196k+188$ is a square.

THEN,

$\displaystyle 196k-7$ is a square.

For some reason I cannot continue. I am hoping someone will see this and tell me the next step. - Aug 24th 2006, 09:15 PMCaptainBlackQuote:

Originally Posted by**OReilly**

by $\displaystyle 49$ for consecutive integral values of $\displaystyle k$.

RonL - Aug 24th 2006, 09:35 PMCaptainBlackQuote:

Originally Posted by**Quick**

**Im**PerfectHacker agrees with you and he doesn't understand his

argument either :D

It is either so subtle that we can't see the point, or its wrong. I know

where I'm putting my money (no not in that box under the bed).

RonL - Aug 25th 2006, 04:32 AMThePerfectHacker
Look at post #7. I say it is no good.

---

I still do not see why you chose such a method. The quickest would be quadradic reciprocity. - Aug 25th 2006, 05:54 AMGlaysherQuote:

Originally Posted by**ThePerfectHacker**

Need to prove either

statement true for n = k implies statement true for n = k + 1

or

Contropositive

Statememt not true for n = k + 1 implies statement not true for n = k

OReilly showed

Statement true for n = k implies statement not true for n = k + 1 - Aug 25th 2006, 06:34 AMQuickQuote:

Originally Posted by**Glaysher**

- Aug 25th 2006, 08:56 AMGlaysherQuote:

Originally Posted by**Quick**

Usually you are trying to prove something for all natural numbers n

You show that it is true when n = 1

Then you assume it is true for some n = k

It is okay to assume this as we have already proved it for n = 1 so we already know there is at least one possible value for k.

We then try to prove that it also holds for n = k + 1 assuming that it holds for n= k.

Once we have done this we have proved it for all natural numbers n as

True for n = 1 means true for n = 2 means true for n = 3 and so on ...

This method can be adapted to suit several different purposes