Find all integer triples $(m,p,q)$ satisfying \[2^mp^2+1=q^5\] where $m>0$ and both $p$ and $q$ are prime numbers.
Problem
Source: Finland 2013, Problem 5
Tags: modular arithmetic, quadratics, search, number theory, Finland
02.05.2013 21:42
Suppose neither $p$ nor $q$ equals $3$; then $2^mp^2 + 1 \equiv 2 \pmod{3}$ when $m$ is even, and $2^mp^2 + 1 \equiv 0 \pmod{3}$ when $m$ is odd. Since $q \ne 3$, the second case is inadmissible. But if $m$ is even, we can write $m = 2m'$, so that \[(2^{m'}p)^2 + 1 = q^5\] By Mihailescu's Theorem, the only consecutive perfect powers are $8$ and $9$, and $9$ is not a fifth power. Thus either $p = 3$ or $q = 3$. If $q = 3$, we have $2^mp^2 = 3^5 - 1 = 242$. This forces $m = 1, p = 11$ which is indeed a solution. Otherwise, $p = 3$ and we need \[9 \cdot 2^m = q^5 - 1\] By Zsigmondy's Theorem, $q^5 - 1$ must have a prime divisor that $q - 1$ does not. Furthermore, $v_2(q^5 - 1) = v_2(q - 1)$ and \[q - 1 | q^5 - 1 = 9 \cdot 2^m\] so $q - 1 = 2^m$. But then \[9 \cdot 2^m = q^5 - 1 = (q - 1)(q^4 + q^3 + q^2 + q + 1) > 2^m(2^{4m})\] so $2^{4m} < 9$, contradicting $m > 0$. The only solution is therefore $(1, 11, 3)$.
03.05.2013 02:21
Another solution in order to avoid the use of Mihailescu's theorem. The given equation becomes $ 2^mp^2=q^5-1 $ From Zsigmondy's theorem there is a prime divisor $ r $ of $ q^5-1 $ that does not divide $ q-1 $ Because $ 2|q-1, \ r=p $ and $ p\ne 2 $ Now from $ 2^mp^2=(q-1)(q^4+q^3+q^2+q+1) $, $ (p,q-1)=1 $ and $ (q^4+q^3+q^2+q+1,2)=1 $, I have that $ q-1=2^m\Rightarrow q=2^m+1 $ Substituting this to the original equation, expanding and dividing by $ 2^m $ I take $ p^2=2^{4m} +5\cdot2^{3m}+10\cdot2^{2m}+10\cdot2^m+5=A $ If $ m\geq2 $ then $ A\equiv5 (mod 8)\Rightarrow p^2\equiv 5 (mod8) $, contradiction. Thus $ m=1 $ which gives $ p=11$ and $q=3 $.
03.05.2013 09:59
I have a different solution. A lot of modulos seem to work but mine is rather big as compared to the others. Firstly, using Catalan's Conjecture/Mihailescu's Theorem, we get that $m$ must be odd, otherwise $x^2+1=q^5$, which does not have integer solutions. So, $m$ is odd. Taking the equation $\pmod{11}$ (the intuition comes from the fact that $m=1,p=11,q=3$ is a solution), we get $2^mp^2 +1 \equiv 0,-1,1 \pmod{11}$ since $0,1$ and $-1$ are the only fifth power residues $\pmod{11}$. Case I: $2^mp^2+1 \equiv 0 \pmod{11} \implies q=11 \implies 2^m.p^2 =11^5-1 \implies m=1$ $\implies 5|p \implies p=5$, an obvious contradiction. Case II: $2^mp^2+1 \equiv -1 \pmod{11} \implies (2^{\frac{m-1}{2}}p)^2 \equiv -1 \pmod{11}$ but $\left(\frac{-1}{11}\right)=-1$, a contradiction. Case III: $2^mp^2 \equiv 0 \pmod{11} \implies p=11$. Now taking $\pmod{3}$, $2^mp^2+1 \equiv 0 \pmod{3} \implies q=3 \implies m=1$. So, the only solution is $(m,p,q)=(1,11,3)$.
05.05.2013 16:10
Not difficult this equation come to \[ y^4+y^3+y^2+y+1 = x^2 \Longrightarrow \] \[ y<4 \] here, $x, y$ - positive integers. Because, if $y\ge 4$ then $ (2y^2+y)^2<(2x)^2<(2y^2+y+1)^2 .$ Hence, from $q=y$ is odd we get $q=3$. Answer: $ (1, 11, 3) $
05.05.2013 17:29
v_Enhance wrote: Find all integer triples $(m,p,q)$ satisfying \[2^mp^2+1=q^5\] where $m>0$ and both $p$ and $q$ are prime numbers. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2775826&sid=ad097457b2837a7cd8257f0389527228#p2775826
22.12.2013 23:01
v_Enhance wrote: Find all integer triples $(m,p,q)$ satisfying \[2^mp^2+1=q^5\] where $m>0$ and both $p$ and $q$ are prime numbers. Moving $1$ to the other side and factoring we obtain $2^mp^2=(q-1)(\frac{q^5-1}{q-1})$. Now note that $\gcd(q-1, \frac{q^5-1}{q-1})\mid 5$. There are now 2 cases to consider. Case 1: $\gcd(q-1, \frac{q^5-1}{q-1})=1$. $\star$ Case 1.1: $p=2$. In this case $2^{m+2}=(q-1)\left(\frac{q^5-1}{q-1}\right)$. Clearly $q^5-1>(q-1)^2$, so $q-1=1\Rightarrow q=2$, yielding $\frac{q^5-1}{q-1}=31$, which is absurd since $2^{m+2}=31$ has no solutions. $\star$ Case 1.2: $p\not= 2$. We saw previously that $q=2$ does not work because it implies $m=0$. So either $q-1=2^m\wedge \frac{q^5-1}{q-1}=p^2$ or $q-1=p^2\wedge \frac{q^5-1}{q-1}=2^m$. The second case gives $q=p^2+1$. If $p$ is odd then this is a contradiction. Otherwise $p=2$, also a contradiction since $p\not= 2$ in this case. So we only consider the first case. We have $q-1=2^m\Rightarrow q=2^m+1$. Now $2^m\equiv (2,1)\pmod 3$ when $m$ is (odd, even) respectively. Therefore if $m$ is odd and greater than $1$ then $q$ is a multiple of $3$ greater than $3$, a contradiction to $q\in \mathbb{P}$. Hence if $m$ is odd then $m=1$, and we see that this works giving $(p,q)=(11,3)$. Now consider when $m$ is even. We get $q=4^k+1$ with $k\ge 1$. Hence $p^2=(4^k+1)^4+(4^k+1)^3+(4^k+1)^2+(4^k+1)+1$. If $k\ge 2$ then this yields $p^2\equiv 5\pmod 8$, a contradiction since $5$ is not a quadratic residue $\mod 8$. So $k=1$ is the only possible solution here. But then $p^2=\frac{5^5-1}{4}=781$ which is a multiple of $11$ (because 1-8+7=0). So since $p$ is prime we must have $p=11$. But $121\not= 781$. Case 2: $\gcd(q-1, \frac{q^5-1}{q-1})=5$. Clearly we must have $p=5$ giving $q-1=5\Rightarrow q=6$, a contradiction. We conclude from our search that $\boxed{(m,p,q)=(1,11,3)}$ is the only solution.
04.02.2014 06:48
$2^m p^2+1=q^5$ we know $q$ is odd. Than $q^5-1=2^mp^2$ than $(q-1)(q^4+q^3+q^2+q+1)=2^mp^2$, $q$ is odd $\Rightarrow$ $q-1=2^mx$ $\Rightarrow$ in case $p=2$, $\emptyset$, thus $p$ is odd. $\Rightarrow$ $x(q^4+q^3+q^2+q+1)=p^2$, $x<q^4+q^3+q^2+q+1$ and $p$ is prime, so $x=1$ and $p^2=q^4+q^3+q^2+q+1$ $\Rightarrow$ $q-1=2^m$ $\Rightarrow$ $q=2^m+1$ $\Leftrightarrow$ $q(q+1)(q^2+1)+1=p^2$. Assume that $p=2p_1+1$ $\Rightarrow$ $(2^m+1)(2^m+2)(2^{2m}+2^{m+1}+2)=4p_1(p_1+1)$. $(2^m+1)(2^{m-1}+1)(2^{2m-1}+2^m+1)=p_1(p_1+1)$. In case $m>1$, $(2^m+1)(2^{m-1}+1)(2^{2m-1}+2^m+1)$ is odd and $p_1(p_1+1)$ is even $\Rightarrow$ Contradiction! $\Rightarrow$ $m=1$ $\Rightarrow$ $q=3$ and $p=11$ Answer: $(p;q;m)=(11;3;1)$. Done!
10.02.2014 19:46
There is no need complicated proofs.We know that $q^4+q^3+q^2+q+1$ is odd and $q^4+q^3+q^2+q+1>q-1$ so $q-1=2^m$ and rest is given.
22.02.2014 07:26
We get $ p^2 - 1=q^4 +q^3 +q^2 +q $ . If $ p $ $ \ge 5 $ ,then $ p^2 - 1 $ is divisibly $ 24 $ we get $ q=3 $ or $ q=2 $, we have $ (p,q,m)=(11,3,1) $. If $ p $ $ \leq 3 $ then no solutions.
25.04.2015 19:08
hyperbolictangent wrote: Suppose neither $p$ nor $q$ equals $3$; then $2^mp^2 + 1 \equiv 2 \pmod{3}$ when $m$ is even, and $2^mp^2 + 1 \equiv 0 \pmod{3}$ when $m$ is odd. Since $q \ne 3$, the second case is inadmissible. But if $m$ is even, we can write $m = 2m'$, so that \[(2^{m'}p)^2 + 1 = q^5\] By Mihailescu's Theorem, the only consecutive perfect powers are $8$ and $9$, and $9$ is not a fifth power. Thus either $p = 3$ or $q = 3$. If $q = 3$, we have $2^mp^2 = 3^5 - 1 = 242$. This forces $m = 1, p = 11$ which is indeed a solution. Otherwise, $p = 3$ and we need \[9 \cdot 2^m = q^5 - 1\] By Zsigmondy's Theorem, $q^5 - 1$ must have a prime divisor that $q - 1$ does not. Furthermore, $v_2(q^5 - 1) = v_2(q - 1)$ and \[q - 1 | q^5 - 1 = 9 \cdot 2^m\] so $q - 1 = 2^m$. But then \[9 \cdot 2^m = q^5 - 1 = (q - 1)(q^4 + q^3 + q^2 + q + 1) > 2^m(2^{4m})\] so $2^{4m} < 9$, contradicting $m > 0$. The only solution is therefore $(1, 11, 3)$. What is Mihailescu's Theorem???
25.04.2015 19:11
http://lmgtfy.com/?q=What+is+Mihailescu%27s+Theorem%3F%3F%3F
03.06.2015 14:36
Why is this problem here???