Sum of squares of numbers equals product of numbers 
Find number of tuples such that the sum of the squares of the numbers equals the product of the numbers.

I tried it and found some tuples like $(3,3,3)$ satisfying $3^2+3^2+3^2=3\times 3\times 3$  but I don't know the real approach to find all such numbers. Can anyone try it?
 A: A partial answer, just to set the ball rolling on this question. As written in the comments the question is too open for a complete answer. This answer is limited to pairs and triples of numbers.

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*There are no numbers such that $a^2+b^2=ab$ (except the trivial solution $0,0$). A quick way to see this is $ab=a^2+b^2\ge2ab\ge ab$ so $a,b=0$.


*If $a^2+b^2+c^2=abc$ then all three numbers must be divisible by $3$. Otherwise if $a=b=c=\pm1\pmod{3}$ then the LHS is divisible by $3$ but not the RHS. More generally, if $m$ divides any pair, say $a,b$, then it must divide the other, $c$. In fact, let $3m$ be the gcd of $(a,b,c)$, with $a=3mA$, $b=3mB$, $c=3mC$, so $$A^2+B^2+C^2=3mABC$$ In this form, no pair can have common factors (if $p|A,B$ then $p^2|C^2$), so all $A,B,C$ consist of distinct primes; also $m$ is not even.
In fact there are an infinite number of solutions with $C=1$ $$A^2+B^2+1=3AB$$ Take any integer solutions of $x^2-5y^2=-4$ (there are an infinite number of these from the theory of quadratic forms). Then $A=\frac{3y\pm x}{2}$, $B=y$, $C=1$, satisfy the above equation, corresponding to the solutions of the original problem $$(3A)^2+(3B)^2+3^2=(3A)(3B)3$$ Thus there are an infinite number of such solutions. The first few combinations for $(A,B,C)$ are $(1,1,1)$, $(2,1,1)$, $(5,2,1)$, $(13,5,1)$, $(34,13,1)$, $(89,34,1)$, $(233,89,1)$, $(610,233,1)$, etc.
Some other observations:

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*If a prime power $p^k$ divides $A$ say, then it also divides $B^2+C^2=A(3mBC-A)$. The sum of squares theorem implies that $p=2$ or $p=1\pmod{4}$; $p=3\pmod{4}$ cannot divide $A$ only, else $B^2+C^2=0\pmod{p}$.

*If $A$ is even, then $B,C$ are odd; but $B^2+C^2$ is not divisible by $4$, so $A$ is not divisible by $4$. Thus either exactly one of $A,B,C$ is divisible by $2$ or none are.

*For each prime $p=1\pmod{4}$, one can solve the problem $A^2+B^2+p^2=3pAB$ using a similar approach as above. Solve $x^2-(9p^2-4)y^2=-4p^2$; then take $A=\frac{3py\pm x}{2}$, $B=y$, $C=p$.

