Playing with equation in integers

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Playing with equation in integers

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#1 h Sep 2, 2019, 7:30 AM • 2 Y

Y by Adventure10, lian_the_noob12

Given the equation
$a^b\cdot b^c=c^a$ in positive integers $a$ , $b$ , and $c$ .
(i) Prove that any prime divisor of $a$ divides $b$ as well.
(ii) Solve the equation under the assumption $b\ge a$ .
(iii) Prove that the equation has infinitely many solutions.

(I. Voronovich)

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#2 h Sep 2, 2019, 9:22 AM • 3 Y

Y by KrysTalk, Adventure10, Mango247

(i) If $a=1,$ then $b=1.$ If $a>1,$ take any $p_{prime} \mid a,$ then $bv_p \left(a \right) + cv_p \left(b \right) = av_p \left(c \right).$ If $p \nmid b,$ then
$bv_p \left(a \right) = av_p \left(c \right) \Rightarrow p^{v_p \left(a \right)} \mid v_p \left(a \right) \Rightarrow v_p \left(a \right) \geqslant 2^{v_p \left(a \right)} \Leftrightarrow v_p \left(a \right) = 0,$ which is impossible. So $p \mid b.$

(ii) If $b \geqslant a,$ then $a^a \mid a^b \mid c^a \implies a \mid c,$ or $c = ka \ \left(k \geqslant 1 \right).$ Then $a^b \cdot b^{ka} = \left(ka \right)^a \iff a^{b-a} \cdot b^{ka} = k^a \Rightarrow k^a \geqslant b^{ka} \Leftrightarrow k \geqslant b^{k},$ So $b=1, \ a = 1, \ c = 1.$

(iii) Take any positive integer $x,$ and $a=c=x^x, \ b = x^{x-1}.$

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#3 h Jan 20, 2020, 11:10 AM • 1 Y

Y by Adventure10

Doxuantrong wrote:

(i) If $a=1,$ then $b=1.$ If $a>1,$ take any $p_{prime} \mid a,$ then $bv_p \left(a \right) + cv_p \left(b \right) = av_p \left(c \right).$ If $p \nmid b,$ then
$bv_p \left(a \right) = av_p \left(c \right) \Rightarrow p^{v_p \left(a \right)} \mid v_p \left(a \right) \Rightarrow v_p \left(a \right) \geqslant 2^{v_p \left(a \right)} \Leftrightarrow v_p \left(a \right) = 0,$ which is impossible. So $p \mid b.$

(ii) If $b \geqslant a,$ then $a^a \mid a^b \mid c^a \implies a \mid c,$ or $c = ka \ \left(k \geqslant 1 \right).$ Then $a^b \cdot b^{ka} = \left(ka \right)^a \iff a^{b-a} \cdot b^{ka} = k^a \Rightarrow k^a \geqslant b^{ka} \Leftrightarrow k \geqslant b^{k},$ So $b=1, \ a = 1, \ c = 1.$

(iii) Take any positive integer $x,$ and $a=c=x^x, \ b = x^{x-1}.$

Can you please explain the first part, from where did you get the equations?

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Cycle

#4 h Jan 20, 2020, 12:25 PM • 2 Y

Y by brokendiamond, Adventure10

i) Suppose $p\mid a$ where $p$ is prime. Since $c^a$ is a multiple of $p$ , so is $c$ , so $p\mid c$ . Then if $p^m\Vert a$ and $p^n\Vert c$ , write $a=p^mx$ and $c=p^ny$ for some $x,y$ not divisible by $p$ . The equation becomes $(p^mx)^b\cdot b^{(p^ny)}=(p^ny)^{(p^mx)}$ and if $p\nmid b$ , equating the powers of $p$ on both sides gives $mb=np^mx$ . But then $p\mid b$ , contradiction. So we must have $p\mid b$ .

ii) If $b\geq a$ , then we have $a^{a+c}=a^a\cdot a^c\leq a^b\cdot b^c=c^a\implies a^{a+c}\leq c^a$ Either $a=c$ , $a>c$ , or $a<c$ . Note that if one of $a,b,c$ is $1$ , it's fairly straightforward to show that we must have $a=b=c=1$ .

If $a=c$ , then we obtain $a=c=1$ which results in a solution $(a,b,c)=(1,1,1)$ .

If $a>c$ , then $a^{2c}<a^{a+c}\leq c^a <a^a$ which is a clear contradiction unless $a=1$ .

If $c>a$ , then we show that $c$ is unbounded unless $c=1$ . Note that $(a^2)^{a}\leq a^{a+c}\leq c^a$ so $a^2\leq c$ . Now if $a^n\leq c$ for some integer $n$ , then $a^{(1+a^{n-1})^{a}} \leq a^{a+c}\leq c^a\implies a^{1+a^{n-1}}\leq c$ and $n<1+a^{n-1}$ unless $a=1$ . So if $a\neq 1$ , starting from $n=2$ , we can repeatedly establish larger and larger lower bounds on $c$ to derive a contradiction. So in this case, we also have $a=b=c=1$ .

In all cases, we conclude that $a=b=c=1$ so the only solution when $b\geq a$ is $a=b=c=1$ .

iii) To show that infinitely many solutions exist, take the family $(a,b,c)=(t^t,t^{t-1},t^t)$ for any positive integer $t$ which is easily shown to work.

This post has been edited 1 time. Last edited by Cycle, Jan 20, 2020, 12:26 PM

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#5 h Sep 6, 2023, 11:32 PM

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(iii):
Construction: $(a,b,c)=(x^x, x^{x-1}, x^x).$
Note, that with this construction, there are infinite, positive integer, values of $x,$ hence, this equation has infinite solutions.

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