# Understand the problem

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# Start with hints

**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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