Some Standard Diophantines

Number Theory Level 3

Find the number of ordered pairs of non-negative integers $(x,y)$ satisfying $x^2+1=y^{2018}.$

The answer is 1.

1 solution

Áron Bán-Szabó
Jun 3, 2018

Note that $y^{2018}=(y^{1009})^2$ . Let $z=y^{1009}$ . Then $x^2+1=z^2$ . Since the squares are $0,1,4,9,\dots$ , the only possible solution where the difference of consecutive squares is 1, is $(x=0, y=1)$ .

Some Standard Diophantines

The answer is 1.

1 solution

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