Library / Macedonian Junior Mathematical Olympiad /6 of 7

Macedonian Junior Mathematical Olympiad

Number theory Difficulty 4.2 AIME Prove it North Macedonia

Solve the equation $x^2 + y^4 + 1 = 6^z$ in the set of integers.

Solution

It is obvious that $z \geq 0$ . If $z \geq 2$ , then $x^2 + y^4 + 1 \equiv 0 \pmod{4}$ , i.e. $x^2 + y^4 \equiv 3 \pmod{4}$ . This is not possible because the remainders of squares of integers after division by $4$ are $0$ or $1$ . According to that, $0 \leq z < 2$ .

If $z = 0$ , then $x = y = 0$ .

If $z=1$ , then $x^2 + y^4 = 5$ , i.e. $(x, y) = \{(2,1), (-2,1), (2,-1), (-2,-1)\}$ .

Therefore $(x, y, z) = \{(0,0,0), (2,1,1), (-2,1,1), (2,-1,1), (-2,-1,1)\}$ are the solutions of the given equation.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.