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
Topic
Algebra
Difficulty Level
3/10

Suggested Book
Challenges and Thrills in Pre College Mathematics

Start with hints

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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