by Guest » Thu Feb 16, 2023 3:28 pm
Let's rearrange the equation a bit:
x^2 - (y+z)x + (y^2+yz+z^2-3) = 0.
This is a quadratic equation in x. In order for x to be an integer, its discriminant must be a perfect square:
(y+z)^2 - 4(y^2+yz+z^2-3) = 4 - 3(y-z)^2.
Let a = y - z. Then we have:
4 - 3a^2 = (y+z)^2 - 4(y^2+yz+z^2-3) = -3y^2-2ay-3a^2-3z^2+12
Simplifying this, we get:
3y^2 + 2ay + 3z^2 = a^2 + 8.
This is a Diophantine equation in y and z. We can treat it as a quadratic in y, and use the quadratic formula to solve for y:
y = (-a ± sqrt(a^2 - 4(3z^2 - a^2 - 8))) / 6.
In order for y to be an integer, the expression under the square root must be a perfect square. Let b = sqrt(3z^2 - a^2 - 8). Then we have:
b^2 + a^2 = 3z^2 - 8.
This is a Diophantine equation in a and z. We can rewrite it as:
(b-a)(b+a) = 3(z^2-3).
Since 3 is prime, we have two cases:
Case 1: 3 divides (b-a) and (b+a). In this case, let b-a = 3c and b+a = 3d, where c and d are integers. Then we have:
2b = 3c+3d,
b^2 + a^2 = 9d^2 - 8,
z^2 = d^2 - c^2 + 3.
The first equation implies that b is divisible by 3. Let b = 3e. Then we have:
e = c+d,
3e^2 + a^2 = 9d^2 - 8,
z^2 = d^2 - c^2 + 3.
The second equation implies that a is odd. Let a = 2f+1. Then we have:
3e^2 + 4f^2 + 4f + 5 = 9d^2,
z^2 = d^2 - c^2 + 3.
Taking both sides of the first equation modulo 3, we see that e and f must have the same parity. If they are both odd, then the left side of the first equation is congruent to 2 modulo 3, which is impossible. Therefore, e and f must be both even. Let e = 2g and f = 2h. Then we have:
3g^2 + 2g + 2h^2 + 3 = 3d^2,
z^2 = d^2 - c^2 + 3.
Taking both sides of the first equation modulo 3, we see that g and h must have the same parity. If they are both odd, then the left side of the first equation is congruent to 2 modulo 3, which is impossible. Therefore, g and h must be both even. Let g = 2i and h = 2j. Then we have:
3i^2 + 2i + 3j^2 + 3 = d