# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]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.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Regional Math Olympiad, 2019 Problem 1[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]3/10

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]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.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

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

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]

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.