Another problem. Let's go!

- ParthKohli

Another problem. Let's go!

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- ParthKohli

Alright, so I think freely when I type - so I'll try to solve this myself. I was able to solve the previous one so I'm hoping that I can do this one just as well.

- ParthKohli

\[(x+1)^p (x-3)^q = x^n + a_1 x^{n-1} + a_2 x^{n-2}+ \cdots + a_n\]Obviously, \(n = p+q\).
We have to find the number of ordered pairs such that \(a_1=a_2\).

- ParthKohli

\[1\le p, q \le 1000\]I'll try to think of this in terms of Vieta's Formulas.

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## More answers

- ParthKohli

So now, here, we have roots \(-1\) and \(3\) with multiplicities \(p\) and \(q\) respectively. Therefore,\[a_1 = - (-1\cdot p + 3 \cdot q) = p - 3q\]This one was simple. Now let's try to find out \(a_2\), which is the sum taken two at a time.

- ParthKohli

Calling @ganeshie8 because I'm feeling quite lonely here.

- ganeshie8

\[a_2 = -3\binom{n}{2}\]
?

- ParthKohli

Really? Let's see.\[\underbrace{-1, \cdots, -1}_{ p~times}, \underbrace{3, \cdots, 3}_{q ~ times}\]In all, there are \(\binom{n}2\) pairs. Out of those, \(\binom{p}2\) pairs contribute 1 each and \(\binom{q}2\) pairs contribute 9 each. The rest contribute -3 each.

- ganeshie8

Ahh right, \(p\ne q\)

- ParthKohli

\[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2 - \binom{p}2 - \binom{q}2\right)\cdot (-3) \]

- ParthKohli

\[= p - 3q\]

- ParthKohli

\[a_2 = \frac{p(p-1) + 9q(q-1)-3\left((p+q)(p+q - 1) - p(p-1) - q(q-1)\right)}{2}\]

- ParthKohli

What an ugly expression... let's see if we can size it down a bit.

- ganeshie8

\(p,q ~\in~\mathbb{N}\) is it ?

- ParthKohli

Yup.

- ParthKohli

Expanded and got\[5p^2 - 3q^2 - 3 p - 3q + 6pq = 0\]This is some kind of a conic section, ain't it?

- ganeshie8

\[8p+24q+1 = (2p-6q+1)^2\]

- ParthKohli

Wow, that looks better. How do we find integral solutions for this?

- ParthKohli

Like it means that \(8p + 24q + 1\) is a perfect square and there aren't many perfect squares in the range 1-1000, but still there are a lot of them and we can't keep track.
You're the number-theory guy here.

- ganeshie8

wait, do we need to find the ordered pairs (p, q) in the range 1-1000 ?

- ParthKohli

Yeah.

- ParthKohli

Oh, so we need to look at perfect squares in the form \(8k + 1\) right?

- ParthKohli

Well, unfortunately, that means all squares of odd numbers. Haha.

- ganeshie8

I think so because the left hand side is 8k+1

- ganeshie8

oh yeah that doesn't help much
@mukushla

- ParthKohli

How did you come up with that form, ganeshie?

- anonymous

Nice question, It's late here guys, I'll come back to this later :) good night!

- ParthKohli

BTW, I found this one on Brilliant too...

- ParthKohli

And good night!

- anonymous

It must be\[p^2 -6pq +9q^2 - 3 p - 3q = 0\]

- anonymous

solve for \(q\)\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6}\]

- ganeshie8

does that mean 16p+1 must be a perfect square

- anonymous

right

- ganeshie8

Nice! there won't be too many perfect squares of this form
is there a nice way to account for divisibility by 6

- ganeshie8

\[j^2 \equiv 1\pmod{16} \implies j\equiv \pm 1, ~\pm 7 \pmod{16}\]

- anonymous

let \(16p+1=n^2\) which gives\[16p=(m-1)(m+1)\]now \(m\) must be an odd number, say \(2k+1\) \[4p=k(k+1)\]how about \(k\)? \(k\) has the forms of \(4l\) or \(4l+3\) only.

- anonymous

* \(16p+1=m^2\)

- ganeshie8

does that mean \(p\) has to be an even triangular number ?

- anonymous

I think yes, let's finish this first

- ganeshie8

\[4p=k(k+1) \implies p =\dfrac{k(k+1)}{2*2}=\dfrac{T_k}{2}\]

- anonymous

aha, counting triangular numbers is easy ha?

- anonymous

we'll work on that too, from mine you get \(p=l(4l+1)\) or \(p=(l+1)(4l+3)\)

- ganeshie8

that looks a lot better! i see where it is going, we work q also similarly right

- anonymous

we have the quadratic formula, we just need to plug p in terms of l to get q

- anonymous

guys dont want to disteurb this meeting after u are done can u explain this to me from point blank point
thanks

- ganeshie8

\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2p+1\pm (8l+7)}{6}\]
?

- anonymous

right

- ganeshie8

subing p also
\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2(l+1)(4l+3)+1\pm (8l+7)}{6}\]

- anonymous

ok we have some pairs of \(p, q\) in terms of \(l\)\[l(4l+1), \frac{l(4l-3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\]

- dan815

|dw:1435058536781:dw|

- dan815

im prolly really behind was this done already

- ganeshie8

right, 4 cases
we can start counting i guess... shouldn't be a problem xD

- anonymous

\[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

- anonymous

this is not good xD

- ganeshie8

ok we have some pairs of \(p, q\) in terms of \(l\)
case1,2
\[l(4l+1), \frac{l(4l-3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\]
case3,4
\[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

- dan815

why did u guys change to L =[

- dan815

|dw:1435059008705:dw|

- ikram002p

i dont know even what u guys trying to solve i'd like to start over

- dan815

6 nice equations i think 3 of 4 are pretty trivial

- ganeshie8

@dan815 equating your expressions for \(a_1\) and \(a_2\), we end up with below quadratic :
\[p^2 -6pq +9q^2 - 3 p - 3q = 0\]

- ganeshie8

then use quadratic formula to isolate \(q\)

- anonymous

@ganeshie8 wait, I found a better method, this is very short

- ganeshie8

ok waiting xD

- anonymous

\[(p-3q)^2=3(p+q)\]right?

- ikram002p

eh i feel like pluto being isolated from this mathematicians council -.-

- ganeshie8

wow ok so \(3\mid (p-3q)^2 \implies 3\mid(p-3q)\)

- anonymous

\(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p-3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;-)

- dan815

there are just 2?

- dan815

oh theres 1 more qeuation possible

- ganeshie8

\[a_1 = p - 3q\tag{1}\]
\[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2 - \binom{p}2 - \binom{q}2\right)\cdot (-3)\tag{2} \]
@dan815 how did u get ur cases from above two equations

- dan815

i kept the cases separate

- ganeshie8

|dw:1435060009607:dw|

- anonymous

guys I have found the final answer as 19, see if this is right or not

- dan815

|dw:1435060528572:dw|

- ganeshie8

19 pairs by both methods!

- ganeshie8

okie

- ikram002p

well that was very exhausted question for me to follow, but indeed unforgettable for a last question >.<

- dan815

I dont see how else u can get it down to only 19

- dan815

lemme think a little more

- dan815

ah omg i found my mistake xD

- ganeshie8

nice :) how many ordered pairs are u getting

- dan815

hey I was afk, i get the same equations i left it there

- dan815

|dw:1435075471169:dw|

- dan815

hmm integer solutions to this eh

- dan815

|dw:1435075773197:dw|

- dan815

|dw:1435075838340:dw|

- dan815

can set a max bound to this now

- ganeshie8

yes :
\(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p-3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;-)

- dan815

how come u got p+q = 3m^2?

- dan815

is it because u know that p+q is an int and we get (p-3q)^2/3 = p+q

- ganeshie8

|dw:1435076243816:dw|

- dan815

right okay
so lets say (p-3q)^2 = 3*m^2
p-3q=+/- 3m

- ganeshie8

\(3\mid (p-3q)^2 \implies 3\mid(p-3q) \implies p-3q = 3m\)

- ganeshie8

|dw:1435076343662:dw|

- dan815

ohhh i see okay ok

- dan815

|dw:1435076425652:dw|