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- Feb 14, 2012
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Find all integer solutions of the system
$a^{a+b}=b^{12}$
$b^{b+a}=a^3$
$a^{a+b}=b^{12}$
$b^{b+a}=a^3$
Bravo, Opalg! And thanks for participating and your elegant piece of solution!\(\displaystyle a^{36} = (a^3)^{12} = (b^{a+b})^{12} = (b^{12})^{a+b} = a^{(a+b)^2}\).
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 $(a,b)= (9,-3).$
another solutions will be a=b=1\(\displaystyle a^{36} = (a^3)^{12} = (b^{a+b})^{12} = (b^{12})^{a+b} = a^{(a+b)^2}\).
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 $(a,b)= (9,-3).$
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
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!another solutions will be a=b=1
and a=1 ,b=-1
so (a,b)=(4,2),(9,-3),(1,1)(1,-1)
Find all integer solutions of the system
$a^{a+b}=b^{12}---(1)$
$b^{b+a}=a^3---(2)$