INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

June 11, 2019

A polynomial equation and roots of unity

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Find all polynomials with real coefficients $P(x)$ such that $P(x^2+x+1)$ divides $P(x^3-1)$.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.23.3" 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="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Indian National Mathematical Olympiad 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Problem Solving Strategies by Arthur Engel[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" 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="3.23.3"]Show that, given a zero of P, it is possible to find another zero.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Whenever such a property holds (as in hint 1), it is a good idea to use the extremal principle (that is, looking at extremal elements).[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Among all (possibly complex) zeros of P, consider the one with the maximum absolute value. Show that, for most choices of P, it is possible to construct another zero with larger absolute value. As that is absurd, a large number of options for P are eliminated.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

As P(x^3-1) is divisible by P(x^2+x+1), there exists a polynomial Q(x) such that P(x^2+x+1)Q(x)=P(x^3-1). Let z be a zero of P. If x^2+x+1=z then x=\frac{-1\pm\sqrt{1+4(z-1)}}{2}, i.e. x^3-1= (x-1)(x^2+x+1)=z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right). The given equation means that z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right) are also zeroes of P. Claim One of the two numbers \left|\frac{-3\pm\sqrt{4z-3}}{2}\right| is greater than 1.   Proof Suppose that they are both \le 1. Then we have    2\ge |-3+\sqrt{4z-3}|   2\ge |-3-\sqrt{4z-3}| Adding these two inequalities we get 4\ge |-3+\sqrt{4z-3}|+|-3-\sqrt{4z-3}|\ge |(-3+\sqrt{4z-3})+(-3-\sqrt{4z-3})|=6 which is absurd. Hence at least one of them has to be greater than 1.     The claim means that, if z is non-zero, then at least one of the two zeroes z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right)  has absolute value greater than |z|.

Now let z=z_0, the zero of P with the largest absolute value. The procedure mentioned above can be used to construct a zero with absolute value bigger than |z_0|, which is absurd unless z_0=0. As z_0 is largest among the zeroes in absolute value, this means that P does not have any nontrivial zeroes. Hence P(x)=ax^n for some a\in\mathbb{R} and n\in\mathbb{N}.

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Watch the video (Coming Soon)

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Math Olympiad Program" url="https://www.cheenta.com/matholympiad/" url_new_window="on" image="https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://www.cheenta.com/matholympiad/" link_option_url_new_window="on"]

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com