Maximum volume of a parallelepiped

Question

The diagonal(space) connecting the 2 opposite vertices of a rectangular parallelepiped is $\sqrt{73}$. Prove that if the squares of the edges of the parallelepiped are integers, then its volume can't exceed 120 units

I tried using AM-GM and I did get the answer. However, I was wondering how we could utilize the fact that squares of all sides are integers here. Any other solutions for this?

P.S.- This was given as a combinatorial problem, hence any proofs involving the same would be appreciated

Answer 1

If the edges are $a,b,c$, then: $$a^2+b^2+c^2=73.$$ Using AM-GM: $$a^2+b^2+c^2\ge 3\sqrt[3]{(abc)^2} \Rightarrow abc\le \left(\frac{73}{3}\right)^{\frac32}\approx 120.03,$$ equality occurs for $a=b=c=\sqrt{\frac{73}{3}}\approx 4.93$.

However, the equation does not have a solution over positive integers, because the possible square numbers are: $$a^2,b^2,c^2=\{1,4,9,16,25,36,49,64\}.$$ The sum of any triple numbers is not equal to $73$.

There are two cases: 1) one number is odd; 2) all three are odd.

Case 1:

$a=\{1,9,25,49\}$. Then $b,c\equiv 0\pmod 4$. No solution.

Case 2:

$a,b,c=\{1,9,25,49\}$. No solution.

Question: Is the original problem stated correctly?

Answer 2

The maximum volume will be closest to a cube. $a^2+b^2+c^2=73$ where $\frac{73}{3} = 24.333$. So $24,24, 25$ will be the closest set of integer values for the square of the sides to yield a maximum volume. $\sqrt(24)*\sqrt(24)*\sqrt(25) = 120$. Hence the volume cannot be greater than $120$.