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

[anemone]

Find all integer solutions of the system

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

$b^{b+a}=a^3$

 

[Opalg]

[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]

 

[anemone]

[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!:)

 

[Albert1]

[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)

 

[Opalg]

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)

 

[Albert1]

if $ a=b=0 $

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

of course we delete a=b=0 as a set of solution

 

[anemone]

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! :(

 

[Albert1]

Find all integer solutions of the system

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

$b^{b+a}=a^3---(2)$

Spoiler

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 :$(a,b)=(4,2),(9,-3),(1,1),(1,-1)$