Suppose that $a, b, x$ are positive integers such that \[x^{a+b}=a^bb\] Prove that $a=x$ and $b=x^x$.
Problem
Source: Iran Third Round MO 1997, Exam 3, P1
Tags: number theory proposed, number theory
mathmdmb
30.07.2010 19:47
Its from Iran,1998 It is obvious that $b\ge a$ and $a,b$ both are perfect powers of $x$.Now $x|a\implies a\ge x\implies a^bb\ge x^bb\implies x^a\ge b$ Let $a=x^k,b=x^l,$ setting we find $x^l+x^k=kx^l+l\implies x^k|l\implies l\ge a \implies b\ge x^a$.So $b=x^a,$then $a=x,b=x^x$
enigmation
30.07.2010 21:25
but why do u say that is obvious ?!! u can't only if a^b and b ar copime
Goutham
31.07.2010 04:32
mathmdmb wrote: It is obvious that $b\ge a$ and $a,b$ both are perfect powers of $x$ Why does it have to hold good? @enigmation pointed out, they are not coprime. mathmdmb wrote: .Now $x|a\implies a\ge x$ Can you say how you got $x|a$? I don't think we can say this unless $x$ is squarefree.
mathmdmb
31.07.2010 11:39
$L.H.S$ is divisible by $x,$ infact a power of $x$.So.If you still dont say it obvious,then Let $a=x^ka_1,b=x^lb_1$ where $k,l$ non-negative integers and $a_1,b_1$ non-divisible by $x$.Now set this in the original equation and we find (still) $L.H.S$ divisible by $x$ where $R.H.S$ is not.Then $a_1=b_1=1$
mathmdmb
31.07.2010 14:44
Let me explain more: Take any arbitrary prime divisor of $x$,say $p.$Now Let, $p^t||a,$ and $p^{a+b}||x^{a+b}$ and $a+b>t$.So $p|b$ too.Thus $x|a,b$.Never is $gcd(a,b)=1$.
tuandokim
31.07.2010 18:02
gouthamphilomath wrote: mathmdmb wrote: It is obvious that $b\ge a$ and $a,b$ both are perfect powers of $x$ Why does it have to hold good? @enigmation pointed out, they are not coprime. mathmdmb wrote: .Now $x|a\implies a\ge x$ Can you say how you got $x|a$? I don't think we can say this unless $x$ is squarefree. It's well-known that: $a^x\ge ax$ for every $a\in N>1$ and $x\in N- {0} $ let $x=\alpha*x_{1},a=\alpha*a_{1}$ where $(a_{1},x_{1})=1$ so $\alpha^{a+b}*x_{1}^{a+b}=\alpha^b*x_{1}^b*b$ so $x_{1}^{a+b} | b$,from the first lemma we got $x_{1}=1$ so x|a mathmdmb wrote: Let me explain more: Take any arbitrary prime divisor of $x$,say $p.$Now Let, $p^t||a,$ and $p^{a+b}||x^{a+b}$ and $a+b>t$.So $p|b$ too.Thus $x|a,b$.Never is $gcd(a,b)=1$. how do u know that p||x??? mathmdmb wrote: $L.H.S$ is divisible by $x,$ infact a power of $x$.So.If you still dont say it obvious,then Let $a=x^ka_1,b=x^lb_1$ where $k,l$ non-negative integers and $a_1,b_1$ non-divisible by $x$.Now set this in the original equation and we find (still) $L.H.S$ divisible by $x$ where $R.H.S$ is not.Then $a_1=b_1=1$ i'm not sure you can get x|a from that.???
mathmdmb
01.08.2010 13:38
My language is completely faulty here.Sorry for I can't express it in a good way.I will be grateful if anybody can explain this (in any way)
Amir Hossein
10.12.2010 12:54
As I was searching for a solution to this problem for a long while, I am posting this. If found this solution in Titu Andreescu's book, Contests Around the World 1998-1999. IMHO, it is not a nice solution. I will appreciate other solutions
Solution: If $x = 1$, then $a = b = 1$ and we are done. So we may assume $x > 1$. Write $x = \prod_{i=1}^n p_i^{\gamma_i},$ where the $p_i$ are the distinct prime factors of $x.$ Since a and b divide $x^{a+b},$ we have $a =\prod p_i^{\alpha_i}$ and $b =\prod p_i^{\beta_i}$ for some non-negative integers $\alpha_i, \beta_i.$
First suppose that some $\beta_i$ is zero, that is, $p_i$ does not divide $b.$ Then the given equation implies that $\gamma_i(a+b) = \alpha_i b,$ so $(\alpha_i - \gamma_i)b=a \gamma_i .$ Now $p_i^{\alpha_i}$ divides $a$ but is coprime to $b$, so $p_i^{\alpha_i}$ divides $\alpha_i - \gamma_i.$ But $p_i^{\alpha_i} > \alpha_i$ for $\alpha_i >0,$ contradiction. We conclude that $\beta_i >0.$
Now from the fact that
\[\gamma_i(a+b) = \beta_i + b \alpha_i \]
and the fact that $p^{\beta_i}$ does not divide $\beta_i$ (again for size reasons), we deduce that $p^{\beta_i}$ also does not divide $a$, that is, $\alpha_i <\beta_i$ for all $i$ and so $a$ divides $b$. Moreover, the equation above implies that $a$ divides $\beta_i,$ so we may write $b = c^a$ with $c \geq 2$ a positive integer.
Write $x/a = p/q$ in lowest terms (so $\gcd(p,q) = 1$). Then the original equation becomes $x^a p^b = b q^b.$ Now $p^b$ must divide $b$, which can only occur if $p = 1$. That is, $x$ divides $a.$
If $x \neq a$, then there exists $i$ with $\alpha_i \geq \gamma_i + 1$, so
\[\gamma_i(a+b) = \beta_i + b \alpha_i \geq (\gamma_i +1) b\]
and so $\gamma_i a > b.$ On the other hand, $a$ is divisible by $p^{\gamma_i}$ , so in particular $a \geq \gamma_i.$ Thus $a^2 > b = c^a,$ or $\sqrt c < a^{1/a};$ However $a^{1/a} < \sqrt 2$ for $a \geq 5,$ so this can only hold for $c = 2$ and $a = 3$, in which case $b = 8$ is not divisible by $a$, contrary to our earlier observation. Thus $x = a$, and from the original equation we get $b = x^x$, as desired.
MATH1945
15.10.2018 14:15
Sorry for the long bump, but does anyone have other solutions than Titu's?
laikhanhhoang_3011
04.06.2021 06:33
MATH1945 wrote: Sorry for the long bump, but does anyone have other solutions than Titu's? i think it is the shortest solution