Problems & Puzzles: Puzzles

Puzzle 351. E796 This week I propose the following puzzle: Find  positive integers, x,y,x such that: x>y>z x^2+y^2-z^2 = a^2 x^2+z^2-y^2 = b^2 y^2+z^2-x^2 = c^2 Perhaps some of you know that L. Euler provided (1781?) four methods in order to produce solutions to the above problem, and also gave specific examples for each of his methods. See the E796 pdf  document. For some non-obvious reason (to me) it happened that the smallest non-trivial solution is a prime solution, that is to say, the triplet (x, y, z) is formed only by prime numbers: (x,y,z)=(269, 241, 149) → (a,b,c)=(329,191,89) (primes in red) Question: Find another prime solution or demonstrate that it does not exist.

---

The second prime solution was found by Jaroslaw Wroblewski.

***

 Wroblewski wrote:

Here is the second prime solution to your Puzzle 351. E796:

 

(x,y,z)=(70201, 64849, 27529) -> (a,b,c)=(91519,38479,5921)

If I am not mistaken, there are no more solutions with x<2,000,000. The method I used allows me to cover x's up to N in time roughly proportional to N. I used parametrizations of 3-term arithmetic progressions of squares of odd integers with no common divisor:

(p+q)^2-2p^2 p^2+q^2 (p+q)^2-2q^2 (raise those to 2nd power)

with p>q and p,q of different parity.

I have checked all solutions to r1^2-2q1^2=r2^2-2q2 (generated by brute
force) with r1,r2<2000 and q1<r1/2, q2<r2/2. I know this is very vague, but at least it should give you general impression of what I have done.

***

 

---