let b and c be integers and let p and q be distinct primes. suppose p^2 + bpq + cq^2 = 0. define a sequence a base n = n^2 + bn +c . show that a base n is not = 0 for all integers n >or = 1.

- anonymous

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- schrodinger

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- Loser66

I can use induction to prove it, but not relate to the condition at all. why do they give out those conditions?

- Loser66

@oldrin.bataku

- anonymous

$$p^2+bpq+cq^2=0\\a_n=n^2+bn+c\ne0\,\forall n\ge1$$

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## More answers

- anonymous

Suppose \(a_n=0\):$$n^2+bn+c=0$$Multiply throughout by \(q^2\):$$n^2q^2+bnq^2+cq^2=0$$

- anonymous

From our first equation, we know $$cq^2=-p^2-bpq$$... so substitute:$$n^2q^2+bnq^2-p^2-bpq=0\\n^2q^2-p^2+bq(nq-p)=0\\(nq-p)(nq+p)+bq(nq-p)=0\\(nq-p)(nq+p+bq)=0\\(nq-p)((n+b)q+p)=0\\\implies nq=p,(n+b)q=p$$... but we know \(p,q\) are distinct primes so we've reached a contradiction.

- Loser66

thanks a lot @oldrin.bataku

- anonymous

Thanks. I have three days trying to figure it out.

- Loser66

It makes perfect sense!!

- anonymous

There has got to be an easier way of doing that though... :-p

- anonymous

I have another question.

- anonymous

My first attempt I tried doing something like this:$$p^2+bpq+cq^2=0\\(p/q)^2+b(p/q)+c=0$$... since \(p,q\) are distinct primes it follows that \(p/q\) is not an integer and thus we've shown \(n^2+bn+c=0\) requires \(n\) to be a non-integer. Thus all integer \(n\ge1\) satisfy \(a_n\ne0\).

- anonymous

I'm not sure if that's valid though

- anonymous

Here is the other question:
Let a, b, c, d be integers. Suppose that both roots of x^2 +bx+c are primes and that f(x)= x^3 + ax^2 + dx +c has three (not necessarily distinct) roots which are integers. Show that f(1)f(-1)=0. Now, assume in addition that a=0. Show that f(2)f(-2)=0.

- anonymous

@oldrin.bataku

- anonymous

Well, if the roots of \(x^2+bx+c\) are primes, say \(p,q\), we know \(b=-(p+q),c=pq\). Looking at our function, we find:
$$f(x)=x^3+ax^2+dx+c\\a=-(k_0+k_1+k_2)\\d=k_0k_1+k_0k_2+k_1k_2\\c=-k_0k_1k_2$$ for roots \(k_0,k_1,k_2\). We then see a relation between our roots via \(c\):$$pq=-k_0k_1k_2$$Because \(p,q\) are prime it follows that one of our roots \(k\) must be \(\pm1\). Thus we see that either \(f(1),f(-1)\) must be \(0\) and thus \(f(1)f(-1)=0\)

- anonymous

That's all?

- Loser66

wait, there is part two, when a =0. just wait

- anonymous

Presuming additionally that \(a=0\), we find either $$0=\pm1+k_1+k_2$$A little algebra gives us:$$k_1+k_2=\mp1$$Now recall that \(k_1k_2=\mp pq\) so either \(k_1,k_2\) are either primes or the additive inverses of primes and we know they sum to \(\mp1\). We know intuitively this means we're looking for primes with a gap of 1 for \(p,q\) and there is only one pair which satisfies this, \(2,3\). Without loss of generality let \(p=2,q=3\) and we therefore know that one of \(k_1,k_2\) must be negative so let \(k_2\) be negative and \(k_1\) positive, and we consider the possibilities:$$3+-2=1\\2+-3=-1$$ so either \(-2,2\) are roots and we find \(f(-2)f(2)=0\)

- anonymous

Thanks.

- anonymous

I have a final question.
Let p be a prime number larger than 3. Prove that p^2 -1 is divisible by 6.

- anonymous

Recognize \(p^2-1=(p+1)(p-1)\). Recognize that since \(p\) is a prime and \(p>3\) we know \(p\) is odd so \(p+1,p-1\) are even and hence \(p^2-1\) is divisible by \(2\).

- anonymous

Now to prove it's divisible by \(3\)... we also know \(p\) cannot be divisible by \(3\) and thus \(p=3k+1\) or \(p=3k+2\) for some \(k\). Since \(p-1,p+1\) we know that, presuming \(p=3k+1\), we end up with \(p-1=3k\) and therefore \(p^2-1\) is divisible by \(3\).
If we presume \(p=3k+2\), however, \(p+1=3k+3\) and therefore \(p^2-1\) is divisible by \(3\).

- Loser66

How can you put it in neat like that?? admire!!

- anonymous

Genius!

- anonymous

Sorry my friend got a question.
Let n be any positive integer. Show that there exists a set consisting of n consecutive integers which are not prime.

- anonymous

Consider the set of integers \(\{(n+1)!+2,(n+1)!+3,(n+1)!+4,\dots,(n+1)!+n+1\}\). None of them are prime, and all have trivial divisors.

- anonymous

How do I prove it?

- anonymous

I dunno, I suppose use induction? It should be easy whichever way. \((n+1)!=(n+1)\cdot n\cdot (n-1)\cdots1\) and so is clearly divisible by \(2,\dots,n+1\). For \((n+1)!+k\) for \(2\le k\le n+1\) it follows then that \(k\ |\ (n+1)!+k\).

- anonymous

Thanks again.

- anonymous

one question. for the question that i previously put ( let p be a prime number larger than 3. prove that p^2-1 is divisible by 6.) you prove that it is divisible by 3. how do i do it for 6? @oldrin.bataku

- Loser66

@mathtutoring22. to me, if a number can be divided by 3 and 2 , it is divided by 6. he proved both them. For example: 2|6 and 3|6 --> 6|6. If there is just one condition, we cannot conclude that. i.e: 3|9 cannot conclude that 6|9.

- anonymous

I already told you. If it is divisible by both 2 and 3 it is also divisible by 6...

- anonymous

Yes. Sorry thanks @oldrin.bataku

- anonymous

I have two more question. @oldrin.bataku

- anonymous

What is greater?
N^n/2 or n!
I have to prove it.

- anonymous

The second question:
Does there exists an positive integer n such that n^2 +n + 41 is not a prime. Yes n=41.
Now show that this is a prime if n<41.

- anonymous

http://www.youtube.com/watch?feature=player_detailpage&v=NsO6nh42oPo

- anonymous

Im sorry. how do i do the second question by induction?

- anonymous

@oldrin.bataku

- anonymous

@mathtutoring22 it's clearly not true \(n<41\) since \(n=40\) yields:$$n^2+n+41=n(n+1)+41=40(41)+41$$

- anonymous

I know. but he wanted us to show it as a math induction problem.

- anonymous

@mathtutoring22 induction is for proving for all \(n\ge k\) but here this does not hold

- anonymous

How do I prove this:
Let a,b be positive integers.
If a^2 + b^2 / 1+ab is an integer, then it is a perfect square.
@oldrin.bataku

- anonymous

hmph... ? is that \(\dfrac{a^2+b^2}{1+ab}\)

- anonymous

Yes

- anonymous

What about the limit of ((1)/(nsin(n)).

- anonymous

With proof.

- anonymous

limit for what?

- anonymous

for the other one:$$\frac{a^2+b^2}{1+ab}=k\text{ for some integer }k\\a^2+b^2=k(1+ab)\\(a+b)^2-2ab=k(1+ab)\\(a+b)^2=k(1+ab)+2ab=k(1+(1+2/k)ab)$$I believe since \(k\) is an integer we may presume \(2/k\) is one as well therefore \(k=1\) or \(k=2\), both of which are squares. Q.E.D.

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