5

I found this competition math problem that I haven't been able to solve.

If $4^x + 4^{-x} = 5$, find $8^x + 8^{-x}$.

Setting $a = 4^x$, we see the problem is equivalent to saying:

If $a + a^{-1} = 5$, find $a^{3/2} + a^{-3/2}$.

So $a$ is the solution to the quadratic equation $a^2 - 5a+ 1 = 0$. Explicitly, $a = \frac{5 + \sqrt{21}}{2}$, which is a unit in the ring integers of $\mathbb Q(\sqrt{21})$. The expression $a+a^{-1}$ is then equal to the trace of $a$. Then $a^{3/2} + a^{-3/2}$ should be another trace.

I would love to see if there is a nice solution to this problem using algebraic number theory. But it should also be possible to solve using elementary methods, since it is a high school competition math problem.

Bill Dubuque
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D_S
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  • $~y^2 + \dfrac{1}{y^2} = \left(y + \dfrac{1}{y}\right)^2 - 2,~$ which implies that (without solving for $~2^x),~$ you can infer the value of $~\displaystyle \left(2^x + 2^{-x}\right).~$ From this, you can infer the value of $\displaystyle 2^{3x} + 2^{-3x} = \left(2^x + 2^{-x}\right)^3 - 3\left(2^x + 2^{-x}\right).$ The point of the question is that it is unnecessary to solve for either $~4^x~$ or $~2^x.$ – user2661923 Oct 21 '23 at 16:09
  • Hint: use symmetric polynomial identities as here or here (this is implicitly done in the answers below). – Bill Dubuque Oct 23 '23 at 01:04

3 Answers3

9

With $a=2^x$ we have $a^2+a^{-2}=5$ which means $a+\frac 1a=\sqrt 7$ or $a-\frac 1a=\sqrt 3$. Then $a^3+a^{-3}=4\sqrt 7$.

abiessu
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A trigonometric solution. Note
$$ 4^x+4^{-x} = e^{x\ln 4}+e^{-x\ln 4} = 2 \cosh(x\ln 4) = 2 \cosh(2x\ln 2), $$ and similarly $$ 8^x+8^{-x} = 2 \cosh(3x\log 2) $$ Writing $t=x\log 2$ we have $$ 4^x+4^{-x} = 2\cosh(2t) = 4\cosh^2(t)-2 \\ 8^x+8^{-x} = 2\cosh(3t) = 8\cosh^3(t)-6\cosh(t) $$ Now we are given $4^x+4^{-x} = 5$, so $$ 5=4\cosh^2(t)-2 \\ \frac{7}{4}=\cosh^2(t) \\ \frac{\sqrt{7}}{2} = \cosh(t) $$ where we assumed $x$ is real so that $\cosh(t) > 0$.
Finally, $$ 8^x+8^{-x} = 8\cosh^3(t)-6\cosh(t) = 8\left(\frac{\sqrt{7}}{2}\right)^3 - 6\left(\frac{\sqrt{7}}{2}\right) =4\sqrt{7} . $$

GEdgar
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2

$(2^x+2^{-x})^2=4^x+4^{-x}+2=7$

$\therefore 8^x+8^{-x}=(2^x+2^{-x})^3-3(2^x+2^{-x})=4\sqrt7$

Dan
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