Diophantine equation

The Diophantine equation x^2 + y^2 + z^2 = 3 x y z have solution with triplets from this series: 1, 2, 5, 13, 29, 34, 89, 169... which are called Markoff numbers.

The solution to this equation where x, y, and z are all primes are: 5, 29 and 433 whose sum is 467 which is also prime. This solution is the smallest we know where all three numbers are prime.

There are other solutions for x, y and z where the sum is prime, but not all the terms are all primes, for example:

x= 1, y = 1597 and z = 4181 are solution to the above Diophantine equation and s = 5779 which is prime, but 2 of the elements are not prime except y = 1597 which is prime.

I'm gonna post a semiformal solution

We can rewrite the equation as a quadratic equation in $x$ , i.e.

$x^2 - x(3yz) + y^2 + z^2=0$

And we can create a solution set $(x,y,z)$ for some $x,y,z$ .

Since we know that any given quadratic equation has two roots (which can sometimes be equal) and the sum of the roots of the quadratic equation above is $3yz$ by Vieta's formulas, then we know that is $x$ is a solution then $3yz-x$ is also a solution. Therefore, if $(x,y,z)$ is a solution set then $(3yz-x, y,z)$ is also a solution set.

Let $(x,y,z) = (1,1,1)$ . Then, we can transform this set by interchanging any two numbers or applying the transformation $(x,y,z) \implies (3yz-x, y,z)$

So , obviously, the first step is $(1,1,1) \implies (2,1,1)$ . Note that retransforming the given set yields $(1,1,1)$ again, so we rearrange the set and retransform, i.e.

$(2,1,1) \implies (1,2,1) \implies (5,2,1)$ .

Since we already have two prime numbers $5,2$ then we just need to rearrange and transform the 1, i.e.

$(5,2,1) \implies (1,2,5) \implies (29,2,5)$ (We cannot transform the 2 since it will yield an odd value, yielding an even value when transforming 1, but since $odd+odd+even = even$ then the sum cannot be prime)

With a prime triplet and a sum $36$ . Uh-oh. Looks like we need to transform again since $36$ is not prime...

$(29,2,5) \implies (2,29,5) \implies (433,29,5)$

Why did we transform the $2$ now? Because $2$ is even, and $even+odd+odd = even$

Since $(433,29,5)$ are all prime, and $433+29+5 = 467$ is also prime, therefore our answer is $\boxed{467}$ .

And, now, I will prove that $467$ is the only possible answer just like how Fermat proved his last theorem. :p

And thus, the proof is complete.

Personally I think that if one does not know Markov numbers or Vieta Jumping that this problem is unreasonably complicated or obscure. However, allow me to take a stab at it: I took a case with many assumptions, and even though this is not a solution, it is a stepping stone: I will choose one of the primes to be 5, since this will make the sum of the squares of the other 2 be divisible by 5. I also assumed, that one of the primes is 2, being the only even prime. Now this gives a third prime as a root: 29. Of course, now s=36 so this isn't a solution. However if we fix two components of the equation: x^2+y^2=k; 3xy=m; leaving us with a quadratic equation: z^2-mz+k=0. Now we see that one of the roots is z, and by Vieta's sum of roots: (z1+z2=m) we have the other root as 3xy-z1. From here we obtain, that if (x, y, z1) is a solution, then (x, y, 3xy-z1) is also a solution. The bad apple in our first solution is the 2, since because of this, our sum is even, and therefore cannot be prime. Now, if we have z1=2, then 3xy-z1=433, which gives us our solution. The idea is that we should be able to find our triplet with one Vieta Jump for the second one gives us back our first triplet (x, y, z). Now, this solution also requires luck and guesswork, but I hope it clears some things up.