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.

Question

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.

loser66 · Answer

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

loser66 · Answer

@oldrin.bataku

anonymous · Answer

$$p^2+bpq+cq^2=0\a_n=n^2+bn+c
e0\,\forall n\ge1$$

anonymous · Answer

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

anonymous · Answer

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 · Answer

thanks a lot @oldrin.bataku

anonymous · Answer

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

loser66 · Answer

It makes perfect sense!!

anonymous · Answer

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

anonymous · Answer

I have another question.

anonymous · Answer

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 · Answer

I'm not sure if that's valid though

anonymous · Answer

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 · Answer

@oldrin.bataku

anonymous · Answer

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 · Answer

That's all?

loser66 · Answer

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

anonymous · Answer

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 · Answer

Thanks.

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

Genius!

anonymous · Answer

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 · Answer

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 · Answer

How do I prove it?

anonymous · Answer

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 · Answer

Thanks again.

anonymous · Answer

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 · Answer

@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 · Answer

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

anonymous · Answer

Yes. Sorry thanks @oldrin.bataku

anonymous · Answer

I have two more question. @oldrin.bataku

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

@oldrin.bataku

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

Yes

anonymous · Answer

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

anonymous · Answer

With proof.

anonymous · Answer

limit for what?

anonymous · Answer

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.