Integer Solutions of $a^{a+b}=b^{12}$ and $b^{b+a}=a^3$

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Discussion Overview

The discussion focuses on finding all integer solutions to the system of equations given by \(a^{a+b}=b^{12}\) and \(b^{b+a}=a^3\). The scope includes mathematical reasoning and exploration of potential solutions.

Discussion Character

  • Mathematical reasoning, Exploratory

Main Points Raised

  • One participant derives that from the first equation, \(a^{36} = (b^{12})^{a+b}\) leads to \((a+b)^2 = 36\), resulting in \(a+b = 6\).
  • From \(a+b = 6\), the second equation simplifies to \(b^6 = a^3\), suggesting \(a = b^2\).
  • Substituting \(a = b^2\) into \(a+b = 6\) leads to the quadratic equation \(a^2 - 13a + 36 = 0\), yielding solutions \((a,b) = (4,2)\) and \((a,b) = (9,-3)\).
  • Another participant mentions additional solutions including \((a,b) = (1,1)\) and \((a,b) = (1,-1)\), while also noting the implications of \(a\) and \(b\) being zero.
  • There is a consensus among participants that \(a = b = 0\) is not a valid solution due to the undefined nature of \(0^0\).
  • Some participants express appreciation for contributions and corrections made during the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the solutions \((4,2)\), \((9,-3)\), \((1,1)\), and \((1,-1)\). However, there is no consensus on the implications of \(a\) and \(b\) being zero, with some participants dismissing it as a solution.

Contextual Notes

The discussion includes various assumptions about the nature of integer solutions and the implications of the equations, but these assumptions are not universally accepted or resolved.

anemone
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Find all integer solutions of the system

$a^{a+b}=b^{12}$

$b^{b+a}=a^3$
 
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[sp]$$a^{36} = (a^3)^{12} = (b^{a+b})^{12} = (b^{12})^{a+b} = a^{(a+b)^2}$$.

Therefore $(a+b)^2 = 36$, $a+b = 6.$ Then the second equation becomes $b^6 = a^3$ and so $a = b^2$. Hence $$a = (6-a)^2 = 36 - 12a + a^2,$$ $$a^2 - 13a + 36 = 0,$$ $$(a-4)(a-9) = 0,$$ giving the solutions $(a,b) = (4,2)$ and $(a,b)= (9,-3).$[/sp]
 
Opalg said:
[sp]$$a^{36} = (a^3)^{12} = (b^{a+b})^{12} = (b^{12})^{a+b} = a^{(a+b)^2}$$.

Therefore $(a+b)^2 = 36$, $a+b = 6.$ Then the second equation becomes $b^6 = a^3$ and so $a = b^2$. Hence $$a = (6-a)^2 = 36 - 12a + a^2,$$ $$a^2 - 13a + 36 = 0,$$ $$(a-4)(a-9) = 0,$$ giving the solutions $(a,b) = (4,2)$ and $(a,b)= (9,-3).$[/sp]

Bravo, Opalg! And thanks for participating and your elegant piece of solution!:)
 
Opalg said:
[sp]$$a^{36} = (a^3)^{12} = (b^{a+b})^{12} = (b^{12})^{a+b} = a^{(a+b)^2}$$.

Therefore $(a+b)^2 = 36$, $a+b = 6.$ Then the second equation becomes $b^6 = a^3$ and so $a = b^2$. Hence $$a = (6-a)^2 = 36 - 12a + a^2,$$ $$a^2 - 13a + 36 = 0,$$ $$(a-4)(a-9) = 0,$$ giving the solutions $(a,b) = (4,2)$ and $(a,b)= (9,-3).$[/sp]
another solutions will be a=b=1
and a=1 ,b=-1
so (a,b)=(4,2),(9,-3),(1,1)(1,-1)
 
Last edited:
Albert said:
another solutions will be a=b=1
or a=1 ,b=-1
We had better not even think about what happens if $a$ and $b$ are both zero. (Evilgrin)
 
if $ a=b=0 $

for $0^0 \,\,undefined, $

of course we delete a=b=0 as a set of solution
 
Albert said:
another solutions will be a=b=1
and a=1 ,b=-1
so (a,b)=(4,2),(9,-3),(1,1)(1,-1)

Cool! Thanks, Albert for catching that...and shame on me because I didn't check with the final solution and was kind of in a haste to reply to Opalg...sorry! :( :o
 
anemone said:
Find all integer solutions of the system

$a^{a+b}=b^{12}---(1)$

$b^{b+a}=a^3---(2)$
taking log function for both sides of (1) and (2)we have :
$\dfrac{log\,a}{log\,b}=4\dfrac{log\,b}{log\,a}$
let $x=\dfrac{log\,a}{log\,b}$
$\therefore x=\dfrac {4}{x}$
$x^2=4 ,\, x=\pm 2$
if $x=\dfrac {log\,a}{log\,b}=2,\, \therefore a=b^2----(3)$
$(1) becomes:a^{a+b}=a^6,\,\, a+b=6---(4)$
from (3)(4) we have (a,b)=(4,2) .and (a,b)=(9,-3)
if $x=\dfrac {log\,a}{log\,b}=-2,\, \therefore a=\dfrac {1}{b^2}$
$\therefore a=b=1 ,\,\, or .\, \, a=1,\,and \,\,\, b=-1$
we conclude :$(a,b)=(4,2),(9,-3),(1,1),(1,-1)$
 

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