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- Feb 14, 2012

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$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

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- Thread starter
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- #1

- Feb 14, 2012

- 3,906

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

- Nov 29, 2013

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Hello.

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

I do not know. At a glance:

[tex]a=1 \ and \ b=1[/tex]

[tex]a=1 \ and \ b=1[/tex]

Regards.

- Mar 31, 2013

- 1,346

Hello.

I do not know. At a glance:

[tex]a=1 \ and \ b=1[/tex]

Regards.

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- Feb 14, 2012

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Suggested solution by J. Chui:

$2^{a^2+b}+2^{a+b^2}=8$

$\sqrt{a}+\sqrt{b}=2$

Suppose that $\sqrt{a}=1+u$ and $\sqrt{b}=1-u$. Then $a+b=2+2u^2 \ge 2$ and $ab=(1-u^2)^2 \le 1$. Thus, by the AM-GM inequality, we have

$8=2^{a^2+b}+2^{a+b^2} \ge 2\sqrt{2^{a^2+b+a+b^2}}\ge 2 \sqrt{2^{(a+b)(a+b+1)-2ab}} \ge 2 \sqrt{2^{2\cdot3-2\cdot1}} \ge 2^3 \ge 8$

with equality iff $a=b$.

Since equality must hold throughtout, $a=b$ and thus the only solution to the system is $(a,b)=(1,1)$.