# very difficult algebra equation

### very difficult algebra equation

Determine all triples (x, y, z) of integers that satisfy the equation
x^2+ y^2+ z^2 - xy - yz - zx = 3.
Note:The correctness of the result must be proved.The correctness of the result must be proved
Kokez

Posts: 7
Joined: Fri Jan 13, 2023 7:00 am
Reputation: 4

### Re: very difficult algebra equation

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
Guest

### Re: very difficult algebra equation

Let's rewrite the given equation as:

x^2 + y^2 + z^2 - (xy + yz + zx) = 3

Now, let's consider the following identity:

(x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2 - xy - yz - zx)

Substituting this identity into the given equation, we get:

(x - y)^2 + (y - z)^2 + (z - x)^2 = 6

Now, we can see that the left-hand side of the equation is a sum of three perfect squares, and the right-hand side is a multiple of 6. Since the only perfect squares that are congruent to 2 modulo 3 are 0 and 1, each of the squares on the left-hand side must be congruent to 2 modulo 3.

However, there is no integer that is congruent to 2 modulo 3, so the equation has no solutions in integers (x, y, z).

Therefore, there are no triples (x, y, z) of integers that satisfy the given equation

apprentus

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Joined: Fri Mar 10, 2023 6:36 am
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