# Understand the problem

Suppose x is a non zero real number such that both $$x^5$$ and $$20 x + \frac{19}{x}$$ are rational numbers. Prove that x is a rational number.

##### Source of the problem
Regional Math Olympiad, 2019 Problem 1
Algebra
3/10

##### Suggested Book
Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

Notice that you can represent higher powers by smaller powers.

Suppose $$20x + \frac{19}{x} = r$$ where r is rational.

Multiply both sides by x to get $$20 x^2 = r \cdot x – 19$$ or $$x^2 = r_1 x + r_2$$ (that is divide by 20 to have r/20 as the coefficient of x and -19/20 as the x free number. both of these are rationals. therefore we name them $$r_1$$ and $$r_2$$ )

Now use the fact that $$x^5$$ is rational.

Suppose $$x^5 = r_3$$ (some rational number).

Then $$x^2 \cdot x^2 \cdot x = r_3$$

But we know $$x^2 = r_1 x + r_2$$

Replacing we have $$(r_1 \cdot x + r_2 )^2 \cdot x = r_3$$

Expanding we have $$r_1^2 x^3 + 2r_1r_2 \cdot x^2 + r_2^2 x = r_3$$

Again we will replace $$x^2$$ to have

$$r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3$$

Expand $$r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3$$

$$r_1^3 x^2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$$

Making one final replacement of x^2 and noting that squaring, adding, multiplying rationals gives rationals we have

$$r_1^3 (r_1 x + r_2) + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$$

$$r_1^4 x + r_1^3r_2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$$

or we have x as a ratio of rationals.

Hence x is rational.

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