Solve the diophantine equation #x-y^4=4# where #x# is a prime?

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Answer 1 · 2017-07-01T11:55:16

A solution is: #x=5#, #y= +-1# or #+-i#

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

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

So, #x-y^4 = 4 -> y^4 = x-4#

Testing prime numbers for #x>4# reveals:

#x=5 -> y^4 = 1#

Given #x=5#, #y= +-1# or #+-i#

NB; This does not prove #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_3y+c_4)# we obtain

#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#