Lagrange's four square_theorem states that every positive integer $N$ can be written as a sum of four squares of integers. At present, let's focus only on positive integer summands; that is, $N=a_1^2+a_2^2+a_3^2+a_4^2$ with $a_i\in\mathbb{N}$.

If now mix-up sums and products, something rather curious happens: the greatest common divisor (gcd) becomes low-primed. I wish to know why.

CLAIM. If $n\geq26$ is an integer, then $$\text{GCD}\,\left\{a_1a_2a_3a_4\,: \, 4n+1=a_1^2+a_2^2+a_3^2+a_4^2 \,\,\, \text{and $a_i\in\mathbb{Z}^{+}$}\right\}=2^b3^c$$ for some $b+c>0$.

For example, \begin{align} 4(26)+1&=7^2+6^2+4^2+2^2=8^2+4^2+4^2+3^2 \\ &=8^2+6^2+2^2+1^2=9^2+4^2+2^2+2^2. \end{align} Therefore, the GCD equals $2^43^1$.

UPDATED.

I now have a more daring claim for the specific powers of $2$ and $3$, showing that even these numbers are low-powered.

CONJECTURE. Denote $a=a(n)$ and $b=b(n)$ the exponents $2^b3^c$ from above. Then, $$b(n)=\begin{cases} 4 \qquad \text{if $n$ is even} \\ 3 \qquad \text{if $n$ is odd}; \end{cases}$$ $$c(n)=\begin{cases} 0 \qquad \text{if $n=3k$} \\ 2 \qquad \text{if $n=3k+1$} \\ 1 \qquad \text{if $n=3k+2$}. \end{cases}$$

"It's easy for number theory to be hard." - anonymous.