where 
is a positive integer. Prove that 
can't be written as a product of 
non-constant polynomials with integer coefficients.
[/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="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Vietnam MO 2014 Problem 2
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Polynomial, Algebra
[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]7/10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskaracharya Prathistan
[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
such that 
.Note that 
and 
for all 
, because they cannot have a real root.Now, we have to make use of the idea that the coefficients are integers. So, some idea of divisibility must come in.The given polynomial is .Try to find out some values of x, for which P(x) is easily determined without n.See, x = 1 and 6 works as \( x^2 - 7x + 6 = (x-1)(x-6) \).
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]So, we saw that x = 1 and x = 6 are good values and we get that P(1) = P(6) = 13.Also, we know that f(1).g(1) | P(1) = 13. So, one of them must be 1.Say, f(1) = 1 = 1 + a + b. Also, f(6) = 36 + 6a - a = 36 + 5a. Now, f(6) | P(6) = 13. So, f(6) = 36 + 5a = 1 or 13. This gives rise to a single case of a = -7; where \( f(x) = x^2 -7x + 7\).But f(x) has real roots implying that P(x) has real roots. Hence contradiction.QED.
[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]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://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%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
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