MHB Find the Minimum Non-Zero Value of A^2+B^2+C^2 with Integer Constraints

[anemone]

Here is this week's POTW:

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Compute the least possible non-zero value of $A^2+B^2+C^2$ such that $A,\,B$, and $C$ are integers satisfying $A\log 16 +B\log 18 +C\log 24 = 0$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Ackbach
2. kaliprasad
3. Olinguito
4. lfdahl

Solution from Ackbach:

Spoiler

We minimize $A^2+B^2+C^2$ subject to $A,B,C\in\mathbb{Z}$ and $A\log(16)+B\log(18)+C\log(24)=0.$ We examine the logarithm relation and reduce it as follows:
\begin{align*}
\log\left(16^A\right)+\log\left(18^B\right)+\log\left(24^C\right)&=0\\
\log\left(16^A18^B24^C\right)&=0\\
16^A18^B24^C&=1\\
\left(2^4\right)^{\!A}\left(2\cdot 3^2\right)^{\!B}\left(2^3\cdot 3\right)^{\!C}&=1\\
2^{4A+B+3C}\cdot 3^{2B+C}&=1\\
4A+B+3C&=0\\
2B+C&=0\\
C&=-2B\\
4A-5B&=0.
\end{align*}
Now we must minimize $A^2+B^2+4B^2=A^2+5B^2$ subject to $4A=5B$. We can plug this into the minimization expression to find that we must minimize
$$\left(\frac54 B\right)^{\!2}+5B^2=\frac{105B^2}{16}.$$
We can't let $B=0,$ or everything is zero, contrary to the problem statement. We need $A,B,C\in\mathbb{Z},$ so $B=1$ doesn't work. The smallest $B$ that allows $A$ and $C$ to be integers is $B=4,$ which forces $A=5,$ and $C=-8$. The smallest value of $A^2+B^2+C^2$ is therefore $16+25+64=105.$ An equivalent solution is $B=-4, A=-5, C=8,$ producing the same minimum.