Solve the diophantine equation $x - y^{4} = 4$ $x-y^4=4$ where $x$ $x$ is a prime?

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Answer 1 · 2017-07-01T11:55:16

A solution is: $x = 5$ $x=5$ , $y = \pm 1$ $y= +-1$ or $\pm i$ $+-i$

A diophantine equation is an equation in which only integer solutions are allowed.

Here we have: $x - y^{4} = 4$ $x-y^4 = 4$ as a diophantine equation, given that $x$ $x$ is a prime number.

So, $x - y^{4} = 4 \to y^{4} = x - 4$ $x-y^4 = 4 -> y^4 = x-4$

Testing prime numbers for $x > 4$ $x>4$ reveals:

$x = 5 \to y^{4} = 1$ $x=5 -> y^4 = 1$

Given $x = 5$ $x=5$ , $y = \pm 1$ $y= +-1$ or $\pm i$ $+-i$

NB; This does not prove $x = 5$ $x=5$ is the only solution.

Answer 2 · 2017-07-01T12:34:48

See below.

Making

$x = y^{4} + 4 = (y^{2} + c_{1} y + c_{2}) (y^{2} + c_{3} y + c_{4})$ $x = y^4+4 = (y^2 + c_1 y + c_2)(y^2+c_3y+c_4)$ we obtain

$c_{1} = - 2, c_{2} = 2, c_{3} = 2, c_{4} = 2$ $c_1 = -2, c_2=2,c_3=2,c_4=2$

then we have the systems

${(y^2-2y+2=1),(y^2+2y+2=x):}$

and

${(y^2-2y+2=x),(y^2+2y+2=1):}$

having as solutions

$x=5,y=1$ and $x=5, y=-1$

Solve the diophantine equation x-y^4=4x−y4=4x-y^4=4 where xxx is a prime?