Solve for integers x, y, z: \begin{align}x + y &= 1

chingli20013a1

chingli20013a1

Answered question

2022-04-22

Solve for integers x, y, z:
x+y=1zx3+y3=1z2.

Answer & Explanation

Payton Cantrell

Payton Cantrell

Beginner2022-04-23Added 15 answers

We have (1z)(x2xy+y2)=1z2.
If z=1,so x+y=0 and we obtain (t,t,1), where t is an integer.
Let z1.
Thus, x2xy+y2=z+1
and x+y=1z,
which gives (1z)23xy=z+1
or 3xy=z23z.
Thus, z is divisible by 3 and
(1z)243(z23z)0
z26z30
312z3+12,
which gives 0z6.

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