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# Number Theory, Korea Junior MO 2015, Problem 7

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]For a polynomial $f(x)$ with integer coefficients and degree no less than $1$, prove that there are infinitely many primes $p$ which satisfies the following. There exists an integer $n$ such that $f(n) \not= 0$ and $|f(n)|$ is a multiple of $p$. [/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="off" _builder_version="4.0" hover_enabled="0"]Korea Junior MO Problem 7 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]8/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="on"]Elementary Number Theory by David Burton [/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"]

# Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][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" hover_enabled="0"]Well, remember the proof that the set of prime numbers is infinite? We started with the assumption that let there be a finite number of prime numbers and then reached a contradiction that there needs to be another extra prime number given that set. Hence, the set of prime numbers is infinite. This problem is also famously known as Schur's Theorem. Observe that the problem can be restated as every nonconstant polynomial p(x) with integer coefficients if S is the set of all nonzero values,
then the set of primes that divide some member of S is infinite. Let us start by assuming that the set is indeed finite. Let $A$ this set of primes $p$ such that $\exists n$ such than $f(n)\ne 0$ and $p|f(n)$. Let |A| be finite.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]If $f(0)=0$ the result is immediate since $p|f(p^n)$ $\forall p$ (just choose $n$ such that $f(p^n)\ne 0$ and so any prime $p\in A$. Now let's take the case when f(0) is non-zero. Let's take $f(x) = a_n.x^n + ... a_1.x + f(0)$.  Now, $f(c.f(0)) = a_n.{c.f(0)}^n + ... a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + ... + a_2.c^2.f(0) + a_1.c + 1 )$. Can you give some appropiate  c to show that another prime must exist?     [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Take c = product of all the primes in A.  Prove that it implies some other prime must exist which is not in A. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Now, $f(c.f(0)) = a_n.{c.f(0)}^n + ... a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + ... + a_2.c^2.f(0) + a_1.c + 1 )$. Observe that if we take c as mentioned then, i.e. c = product of all the primes in A. Then all f(c.f(0)) must be coprime to all the primes in A. Therefore, it must have a prime factor other than those in A. Hence, a contradiction in the finiteness in A. 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"]

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# Connected Program at Cheenta

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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%%" 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"]

# Similar Problems

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