ParthKohli
  • ParthKohli
Another problem. Let's go!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • 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
  • 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
  • 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
  • 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
  • ParthKohli
Calling @ganeshie8 because I'm feeling quite lonely here.
ganeshie8
  • ganeshie8
\[a_2 = -3\binom{n}{2}\] ?
ParthKohli
  • 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
  • ganeshie8
Ahh right, \(p\ne q\)
ParthKohli
  • 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
  • ParthKohli
\[= p - 3q\]
ParthKohli
  • 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
  • ParthKohli
What an ugly expression... let's see if we can size it down a bit.
ganeshie8
  • ganeshie8
\(p,q ~\in~\mathbb{N}\) is it ?
ParthKohli
  • ParthKohli
Yup.
ParthKohli
  • 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
  • ganeshie8
\[8p+24q+1 = (2p-6q+1)^2\]
ParthKohli
  • ParthKohli
Wow, that looks better. How do we find integral solutions for this?
ParthKohli
  • 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
  • ganeshie8
wait, do we need to find the ordered pairs (p, q) in the range 1-1000 ?
ParthKohli
  • ParthKohli
Yeah.
ParthKohli
  • ParthKohli
Oh, so we need to look at perfect squares in the form \(8k + 1\) right?
ParthKohli
  • ParthKohli
Well, unfortunately, that means all squares of odd numbers. Haha.
ganeshie8
  • ganeshie8
I think so because the left hand side is 8k+1
ganeshie8
  • ganeshie8
oh yeah that doesn't help much @mukushla
ParthKohli
  • ParthKohli
How did you come up with that form, ganeshie?
anonymous
  • anonymous
Nice question, It's late here guys, I'll come back to this later :) good night!
ParthKohli
  • ParthKohli
BTW, I found this one on Brilliant too...
ParthKohli
  • ParthKohli
And good night!
anonymous
  • anonymous
It must be\[p^2 -6pq +9q^2 - 3 p - 3q = 0\]
anonymous
  • anonymous
solve for \(q\)\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6}\]
ganeshie8
  • ganeshie8
does that mean 16p+1 must be a perfect square
anonymous
  • anonymous
right
ganeshie8
  • 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
  • ganeshie8
\[j^2 \equiv 1\pmod{16} \implies j\equiv \pm 1, ~\pm 7 \pmod{16}\]
anonymous
  • 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
  • anonymous
* \(16p+1=m^2\)
ganeshie8
  • ganeshie8
does that mean \(p\) has to be an even triangular number ?
anonymous
  • anonymous
I think yes, let's finish this first
ganeshie8
  • ganeshie8
\[4p=k(k+1) \implies p =\dfrac{k(k+1)}{2*2}=\dfrac{T_k}{2}\]
anonymous
  • anonymous
aha, counting triangular numbers is easy ha?
anonymous
  • anonymous
we'll work on that too, from mine you get \(p=l(4l+1)\) or \(p=(l+1)(4l+3)\)
ganeshie8
  • ganeshie8
that looks a lot better! i see where it is going, we work q also similarly right
anonymous
  • anonymous
we have the quadratic formula, we just need to plug p in terms of l to get q
anonymous
  • anonymous
guys dont want to disteurb this meeting after u are done can u explain this to me from point blank point thanks
ganeshie8
  • ganeshie8
\[q=\frac{2p+1 \pm \sqrt{16p+1}}{6} = \dfrac{2p+1\pm (8l+7)}{6}\] ?
anonymous
  • anonymous
right
ganeshie8
  • 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
  • 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
  • dan815
|dw:1435058536781:dw|
dan815
  • dan815
im prolly really behind was this done already
ganeshie8
  • ganeshie8
right, 4 cases we can start counting i guess... shouldn't be a problem xD
anonymous
  • anonymous
\[(l+1)(4l+3), \frac{(l+1)(4l+7)}{3} \\ (l+1)(4l+3), \frac{4l^2}{3}+l\]
anonymous
  • anonymous
this is not good xD
ganeshie8
  • 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
  • dan815
why did u guys change to L =[
dan815
  • dan815
|dw:1435059008705:dw|
ikram002p
  • ikram002p
i dont know even what u guys trying to solve i'd like to start over
dan815
  • dan815
6 nice equations i think 3 of 4 are pretty trivial
ganeshie8
  • 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
  • ganeshie8
then use quadratic formula to isolate \(q\)
anonymous
  • anonymous
@ganeshie8 wait, I found a better method, this is very short
ganeshie8
  • ganeshie8
ok waiting xD
anonymous
  • anonymous
\[(p-3q)^2=3(p+q)\]right?
ikram002p
  • ikram002p
eh i feel like pluto being isolated from this mathematicians council -.-
ganeshie8
  • ganeshie8
wow ok so \(3\mid (p-3q)^2 \implies 3\mid(p-3q)\)
anonymous
  • 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
  • dan815
there are just 2?
dan815
  • dan815
oh theres 1 more qeuation possible
ganeshie8
  • 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
  • dan815
i kept the cases separate
ganeshie8
  • ganeshie8
|dw:1435060009607:dw|
anonymous
  • anonymous
guys I have found the final answer as 19, see if this is right or not
dan815
  • dan815
|dw:1435060528572:dw|
ganeshie8
  • ganeshie8
19 pairs by both methods!
ganeshie8
  • ganeshie8
okie
ikram002p
  • ikram002p
well that was very exhausted question for me to follow, but indeed unforgettable for a last question >.<
dan815
  • dan815
I dont see how else u can get it down to only 19
dan815
  • dan815
lemme think a little more
dan815
  • dan815
ah omg i found my mistake xD
ganeshie8
  • ganeshie8
nice :) how many ordered pairs are u getting
dan815
  • dan815
hey I was afk, i get the same equations i left it there
dan815
  • dan815
|dw:1435075471169:dw|
dan815
  • dan815
hmm integer solutions to this eh
dan815
  • dan815
|dw:1435075773197:dw|
dan815
  • dan815
|dw:1435075838340:dw|
dan815
  • dan815
can set a max bound to this now
ganeshie8
  • 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
  • dan815
how come u got p+q = 3m^2?
dan815
  • dan815
is it because u know that p+q is an int and we get (p-3q)^2/3 = p+q
ganeshie8
  • ganeshie8
|dw:1435076243816:dw|
dan815
  • dan815
right okay so lets say (p-3q)^2 = 3*m^2 p-3q=+/- 3m
ganeshie8
  • ganeshie8
\(3\mid (p-3q)^2 \implies 3\mid(p-3q) \implies p-3q = 3m\)
ganeshie8
  • ganeshie8
|dw:1435076343662:dw|
dan815
  • dan815
ohhh i see okay ok
dan815
  • dan815
|dw:1435076425652:dw|
dan815
  • dan815
that simplfies nicely wow now we just check perfect squares *3 and below 2000
ganeshie8
  • ganeshie8
that is only a necessary condition \(p+q=3m^2\) for \(m \in \mathbb{N}\) and \(p-3q=\pm3 m\) solve for \(p\) and \(q\) in terms of \(m\) ;-)

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