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ParthKohli
 one year ago
Another problem. Let's go!
ParthKohli
 one year ago
Another problem. Let's go!

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Alright, 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
 one year ago
Best ResponseYou've already chosen the best response.5\[(x+1)^p (x3)^q = x^n + a_1 x^{n1} + a_2 x^{n2}+ \cdots + a_n\]Obviously, \(n = p+q\). We have to find the number of ordered pairs such that \(a_1=a_2\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[1\le p, q \le 1000\]I'll try to think of this in terms of Vieta's Formulas.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5So 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
 one year ago
Best ResponseYou've already chosen the best response.5Calling @ganeshie8 because I'm feeling quite lonely here.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[a_2 = 3\binom{n}{2}\] ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Really? 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
 one year ago
Best ResponseYou've already chosen the best response.4Ahh right, \(p\ne q\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2  \binom{p}2  \binom{q}2\right)\cdot (3) \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5\[a_2 = \frac{p(p1) + 9q(q1)3\left((p+q)(p+q  1)  p(p1)  q(q1)\right)}{2}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5What an ugly expression... let's see if we can size it down a bit.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(p,q ~\in~\mathbb{N}\) is it ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Expanded and got\[5p^2  3q^2  3 p  3q + 6pq = 0\]This is some kind of a conic section, ain't it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[8p+24q+1 = (2p6q+1)^2\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Wow, that looks better. How do we find integral solutions for this?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Like it means that \(8p + 24q + 1\) is a perfect square and there aren't many perfect squares in the range 11000, but still there are a lot of them and we can't keep track. You're the numbertheory guy here.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4wait, do we need to find the ordered pairs (p, q) in the range 11000 ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Oh, so we need to look at perfect squares in the form \(8k + 1\) right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Well, unfortunately, that means all squares of odd numbers. Haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think so because the left hand side is 8k+1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4oh yeah that doesn't help much @mukushla

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5How did you come up with that form, ganeshie?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nice question, It's late here guys, I'll come back to this later :) good night!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5BTW, I found this one on Brilliant too...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It must be\[p^2 6pq +9q^2  3 p  3q = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve for \(q\)\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does that mean 16p+1 must be a perfect square

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Nice! there won't be too many perfect squares of this form is there a nice way to account for divisibility by 6

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[j^2 \equiv 1\pmod{16} \implies j\equiv \pm 1, ~\pm 7 \pmod{16}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let \(16p+1=n^2\) which gives\[16p=(m1)(m+1)\]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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does that mean \(p\) has to be an even triangular number ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think yes, let's finish this first

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[4p=k(k+1) \implies p =\dfrac{k(k+1)}{2*2}=\dfrac{T_k}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0aha, counting triangular numbers is easy ha?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we'll work on that too, from mine you get \(p=l(4l+1)\) or \(p=(l+1)(4l+3)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that looks a lot better! i see where it is going, we work q also similarly right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have the quadratic formula, we just need to plug p in terms of l to get q

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0guys dont want to disteurb this meeting after u are done can u explain this to me from point blank point thanks

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2p+1\pm (8l+7)}{6}\] ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4subing p also \[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2(l+1)(4l+3)+1\pm (8l+7)}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok we have some pairs of \(p, q\) in terms of \(l\)\[l(4l+1), \frac{l(4l3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1im prolly really behind was this done already

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4right, 4 cases we can start counting i guess... shouldn't be a problem xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4ok we have some pairs of \(p, q\) in terms of \(l\) case1,2 \[l(4l+1), \frac{l(4l3)}{3} \\ l(4l+1), \frac{(l+1)(4l+1)}{3}\] case3,4 \[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1why did u guys change to L =[

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0i dont know even what u guys trying to solve i'd like to start over

dan815
 one year ago
Best ResponseYou've already chosen the best response.16 nice equations i think 3 of 4 are pretty trivial

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4@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
 one year ago
Best ResponseYou've already chosen the best response.4then use quadratic formula to isolate \(q\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 wait, I found a better method, this is very short

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(p3q)^2=3(p+q)\]right?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0eh i feel like pluto being isolated from this mathematicians council .

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4wow ok so \(3\mid (p3q)^2 \implies 3\mid(p3q)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh theres 1 more qeuation possible

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[a_1 = p  3q\tag{1}\] \[a_2 = \binom{p}2\cdot 1 + \binom{q}2\cdot 9 + \left(\binom{p+q}2  \binom{p}2  \binom{q}2\right)\cdot (3)\tag{2} \] @dan815 how did u get ur cases from above two equations

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i kept the cases separate

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435060009607:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0guys I have found the final answer as 19, see if this is right or not

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.419 pairs by both methods!

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0well that was very exhausted question for me to follow, but indeed unforgettable for a last question >.<

dan815
 one year ago
Best ResponseYou've already chosen the best response.1I dont see how else u can get it down to only 19

dan815
 one year ago
Best ResponseYou've already chosen the best response.1lemme think a little more

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ah omg i found my mistake xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nice :) how many ordered pairs are u getting

dan815
 one year ago
Best ResponseYou've already chosen the best response.1hey I was afk, i get the same equations i left it there

dan815
 one year ago
Best ResponseYou've already chosen the best response.1hmm integer solutions to this eh

dan815
 one year ago
Best ResponseYou've already chosen the best response.1can set a max bound to this now

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4yes : \(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1how come u got p+q = 3m^2?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1is it because u know that p+q is an int and we get (p3q)^2/3 = p+q

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435076243816:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1right okay so lets say (p3q)^2 = 3*m^2 p3q=+/ 3m

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(3\mid (p3q)^2 \implies 3\mid(p3q) \implies p3q = 3m\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1435076343662:dw