What is the smallest number $k$ that is a sum of three squares in two different ways(there does not exist a pair of numbers from different triple which are the same) $$a_0^2+b_0^2+c_0^2=a_1^2+b_1^2+c_1^2=k$$ Such that $a_0b_0c_0=a_1b_1c_1$?
Also, does there exist a solution in two numbers?
Note: All the numbers mentioned are non-zero natural numbers
My Attempt:
I’ve started with the two variable case as it seemed easier, and I got to the equation:
$$a_0^2k-a_0^4=a_1^2k-a_1^4$$
Needing to have two solutions where $a$ and $b$ are distinct for some fixed $k$.
It looks like it implies no solutions but I had not managed to prove it.
I’ve also tried parity arguments to no success.
I've found a pair of triples, $1^2+12^2+10^2=2^2+4^2+15^2 = 245$ and $1\cdot 12 \cdot 10 = 2\cdot 4\cdot 15=120$ but I'm unsure of it's minimality.
Triples under $99$ (excluding equal product considerations) include
\begin{align*} 1^2 + 4^2 + 4^2 = 2^2 + 2^2 + 5^2 = 33\\ 1^2 + 1^2 + 6^2 = 2^2 + 3^2 + 5^2 = 38\\ 1^2 + 2^2 + 6^2 = 3^2 + 4^2 + 4^2 = 41\\ 1^2 + 2^2 + 7^2 = 2^2 + 5^2 + 5^2 = 3^2 + 3^2 + 6^2 = 54\\ 1^2 + 5^2 + 6^2 = 2^2 + 3^2 + 7^2 = 62\\ 1^2 + 1^2 + 8^2 = 1^2 + 4^2 + 7^2 = 66\\ 1^2 + 2^2 + 8^2 = 2^2 + 4^2 + 7^2 = 69\\ 1^2 + 3^2 + 8^2 = 3^2 + 4^2 + 7^2 = 74\\ 2^2 + 3^2 + 8^2 = 4^2 + 5^2 + 6^2 = 77\\ 1^2 + 4^2 + 8^2 = 3^2 + 6^2 + 6^2 = 81\\ 1^2 + 1^2 + 9^2 = 3^2 + 5^2 + 7^2 = 83\\ 1^2 + 2^2 + 9^2 = 1^2 + 6^2 + 7^2 = 5^2 + 5^2 + 6^2 = 86\\ 2^2 + 2^2 + 9^2 = 2^2 + 7^2 + 6^2 = 3^2 + 4^2 + 8^2 = 89\\ 1^2 + 5^2 + 8^2 = 4^2 + 5^2 + 7^2 = 90\\ 2^2 + 3^2 + 9^2 = 3^2 + 2^2 + 9^2 = 3^2 + 6^2 + 7^2 = 94\\ 1^2 + 4^2 + 9^2 = 3^2 + 5^2 + 8^2 = 98\\ 1^2 + 7^2 + 7^2 = 3^2 + 3^2 + 9^2 = 5^2 + 5^2 + 7^2 = 99\\ \end{align*} I put these and all others up to 245 in a spreadsheet with product formulas for each triple. Only those for the sum 245 had more than one triple with the same product.