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- #1

- Feb 14, 2012

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Find all integer solutions of the system

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

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

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

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

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- Thread starter
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- #1

- Feb 14, 2012

- 3,805

Find all integer solutions of the system

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

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

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

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

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- #2

- Feb 7, 2012

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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 $(a,b)= (9,-3).$

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- #3

- Feb 14, 2012

- 3,805

Bravo,

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 $(a,b)= (9,-3).$

- Jan 25, 2013

- 1,225

another solutions will be a=b=1

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 $(a,b)= (9,-3).$

and a=1 ,b=-1

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

Last edited:

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- #5

- Feb 7, 2012

- 2,738

We had better not even think about what happens if $a$ and $b$ are both zero.another solutions will be a=b=1

or a=1 ,b=-1

- Jan 25, 2013

- 1,225

if $ a=b=0 $

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

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

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

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

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- #7

- Feb 14, 2012

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Cool! Thanks,another solutions will be a=b=1

and a=1 ,b=-1

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

- Jan 25, 2013

- 1,225

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