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# 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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