Prime Again?

Number Theory Level 4

{ x 2 + x y + y 2 = 19 y 2 + y z + z 2 = 67 z 2 + z x + x 2 = 79 \large{\begin{cases} x^2 + xy + y^2 &=& 19 \\ y^2 + yz + z^2 &=& 67 \\ z^2 + zx + x^2 &=& 79 \\ \end{cases} }

Considering the system of Diophantine equations above, what is the value of x y + y z + z x xy + yz +zx ?


The answer is 41.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

From the first equation, x 2 + x y + y 2 = 19 x^2 + xy + y^2 = 19 , when we solve for x in terms of y, it can be calculated with a quadratic formula:

x = − y ± y 2 − 4 ( y 2 − 19 ) 2 \dfrac{-y\pm \sqrt{y^2 - 4(y^2 - 19)}}{2} = − y ± 76 − 3 y 2 2 \dfrac{-y\pm \sqrt{76 - 3y^2 }}{2}

Considering the root of the quadratic discriminant, 76 − 3 y 2 76 - 3y^2 > 0. Hence, y y ∈ \in [-5 , 5].

Since this system is of Diophantine equations, the solutions of all variables are integers. Therefore, when testing for y-integers, we will get:

y = -5 → \to x = 5 ± 76 − 3 ( 5 2 ) 2 \dfrac{5\pm \sqrt{76 - 3(5^2)}}{2} = 5 ± 1 2 \dfrac{5\pm 1}{2} = 2 or 3.

y = -4 → \to x = 4 ± 76 − 3 ( 4 2 ) 2 \dfrac{4\pm \sqrt{76 - 3(4^2)}}{2} = 4 ± 28 2 \dfrac{4\pm \sqrt{28}}{2} (not integers)

y = -3 → \to x = 3 ± 76 − 3 ( 3 2 ) 2 \dfrac{3\pm \sqrt{76 - 3(3^2)}}{2} = 3 ± 7 2 \dfrac{3\pm 7}{2} = -2 or 5.

y = -2 → \to x = 2 ± 76 − 3 ( 2 2 ) 2 \dfrac{2\pm \sqrt{76 - 3(2^2)}}{2} = 2 ± 8 2 \dfrac{2\pm 8}{2} = -3 or 5.

y = -1 → \to x = 1 ± 76 − 3 ( 1 2 ) 2 \dfrac{1\pm \sqrt{76 - 3(1^2)}}{2} = 1 ± 73 2 \dfrac{1\pm \sqrt{73}}{2} (not integers)

y = 0 → \to x = 0 ± 76 − 3 ( 0 2 ) 2 \dfrac{0\pm \sqrt{76 - 3(0^2)}}{2} = 76 2 \dfrac{ \sqrt{76}}{2} (not integers)

y = 1 → \to x = − 1 ± 76 − 3 ( 1 2 ) 2 \dfrac{-1\pm \sqrt{76 - 3(1^2)}}{2} = − 1 ± 73 2 \dfrac{-1\pm \sqrt{73}}{2} (not integers)

y = 2 → \to x = − 2 ± 76 − 3 ( 2 2 ) 2 \dfrac{-2\pm \sqrt{76 - 3(2^2)}}{2} = − 2 ± 8 2 \dfrac{-2\pm 8}{2} = 3 or -5.

y = 3 → \to x = − 3 ± 76 − 3 ( 3 2 ) 2 \dfrac{-3\pm \sqrt{76 - 3(3^2)}}{2} = − 3 ± 7 2 \dfrac{-3\pm 7}{2} = 2 or -5.

y = 4 → \to x = − 4 ± 76 − 3 ( 4 2 ) 2 \dfrac{-4\pm \sqrt{76 - 3(4^2)}}{2} = − 4 ± 28 2 \dfrac{-4\pm \sqrt{28}}{2} (not integers)

y = 5 → \to x = − 5 ± 76 − 3 ( 5 2 ) 2 \dfrac{-5\pm \sqrt{76 - 3(5^2)}}{2} = − 5 ± 1 2 \dfrac{-5\pm 1}{2} = -2 or -3.

As a result, the possible pairs of (x , y) include:

{(2 , -5), (2 , 3), (3 , -5), (3 , 2), (5 , -2), (5 , -3), (-2 , -3), (-2 , 5), (-3 , -2), (-3 , 5), (-5 , 2), (-5 , 3)}

Similarly, when we evaluate the second equation, the possible pairs of (y , z) include:

{(2 , 7), (2 , -9), (7 , 2), (7 , -9), (9 , -2), (9 , -7), (-2 , 9), (-2 , -7), (-7 , 9), (-7 , -2), (-9 , 2), (-9 , 7)}

Finally, for the third equation, the possible pairs of (z , x) include:

{(3 , 7), (3 , -10), (7 , 3), (7 , -10), (10 , -3), (10 , -7), (-3 , 10), (-3 , -7), (-7 , 10), (-7 , -3), (-10 , 7), (-10 , 3)}

Now the only applicable triples (x , y , z) corresponding to those integers include: (3 , 2 , 7) or (-3 , -2 , -7).

Either way, the value of x y + y z + z x xy + yz +zx = 6 + 14 + 21 = 41.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that ( 5 21 , 17 21 , − 43 21 ) (\frac{5}{\sqrt{21}}, \frac{17}{\sqrt{21}}, -\frac{43}{\sqrt{21}}) and ( − 5 21 , − 17 21 , 43 21 ) ( -\frac{5}{\sqrt{21}}, -\frac{17}{\sqrt{21}}, \frac{43}{\sqrt{21}}) are also solutions, which lead to x y + x z + y z = − 41 xy + xz + yz = -41 . The problem should say that only integer solutions are desired.

Jon Haussmann - 5 years, 5 months ago

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Sorry, I wrote Diophantine equations before. It was probably erased after edition.

Worranat Pakornrat - 5 years, 5 months ago

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