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November 9, 2019

Number Theory, Ireland MO 2018, Problem 9

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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" box_shadow_style="preset2"]The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$.
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

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Ireland MO 2018, Problem 9 [/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="off"]Excursion in Mathematics by Bhaskaryacharya Prathisthan [/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

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"], wIt is so important to know and use the modulo technqiue at the right time.  We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.  

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Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

  • Cubic(3rd) Power Residue of 9 is {0,1,-1}.
  • 6th Power Residue of 9 is {0,1}
  • Quadratic(2nd Power) Residue of 9 is {0,1,4,7}
We will use these ideas here.   [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$. Note that, $a_{k+1}=a^6+2018$.  Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: \( a_k = 0\) mod 9 Then, the sequence modulo 9 will be  $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2: \( a_k = 1,-1\) mod 9 Then, the sequence modulo 9 will be  $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9  are {0,1,-1}. 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"]

Watch video

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Connected Program at Cheenta

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Similar Problems

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