Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
mukushla
Group Title
Find all triples \((x,m,n)\) of positive integers satisfying the equation \[x^m=2^{2n+1}+2^n+1\]
 one year ago
 one year ago
mukushla Group Title
Find all triples \((x,m,n)\) of positive integers satisfying the equation \[x^m=2^{2n+1}+2^n+1\]
 one year ago
 one year ago

This Question is Closed

mukushla Group TitleBest ResponseYou've already chosen the best response.2
Well note that \(x\) is an odd number. \(m=1 \) is an obvious answer so i want to work on \(m>1 \).............. firstly suppose that \(m\) is an odd number \(m=2k+1 \) where \(k \ge 1 \) we have \[ 2^{n}(2^{n+1}+1)=x^{m}1=(x1)(x^{m1}+...+1)\]\( (x^{m1}+x^{m2}+...+x+1)=(x^{2k}+x^{2k1}...+x+1) \) is an odd number since \(x\) is odd so \(x1\) divides \(2^n\) then \(x1 \ge 2^n\) and \(x\ge 1+2^n\) hence : \[ 2^{2n+1}+2^{n}+1=x^{m}\geq (2^{n}+1)^{3}=2^{3n}+3\times 2^{2n}+3\times 2^{n}+1 \]and this is a contradiction.....so \(m\) is even now let \(m=2k\) then \( 2^{n}(2^{n+1}+1)=x^{2k}1=(x^k1)(x^k+1)=(y1)(y+1) \) and \(y\) is odd. \(y1\) and \(y+1\) are even factors so \( \gcd(y1,y+1)=2 \) exactly one of them divisible by \(4\). Hence \(n ≥ 3\) and one of these factors is divisible by \(2^{n−1}\) but not by \(2^{n} \). So we can write \(y = 2^{n−1}s +t\) , \(s\) odd , \(t= ±1\). Plugging this into the original equation we obtain \(2^n(2^{n+1}+1)=(2^{n−1}s +t)^21=2^{2n−2}s^2+2^n st \) or, equivalently \(2^{n+1}+1=2^{n−2}s^2+ st \) Therefore \(1st=2^{n−2}(s^28) \) For \(t= 1\) this yields \(s^2 − 8 ≤ 0\), so \(s = 1\), which fails to satisfy original equation. For \(t=1\) equation gives \(1+s=2^{n−2}(s^28) \ge 2(s^28)\) so \(2s^2s17 \le 0\) and \( s \le 3\) we know that \(s\) is odd so \(s=3\). put this in the last equation with \(t=1 \) gives \(n=4\). back to the original equation with \(n=4\) u have \[x^m=529=23^2 \] so \(x=23\) and \(m=2\) the only solutions are \( (x,m,n)=(2^{2l+1}+2^l+1, 1,l)\) and \( (23,2,4)\)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.