Number Theory, South Africa 2019, Problem 6

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Understand the problem

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Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$$ 20^m - 10m^2 + 1 = 19^n. $$
[/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"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]South Africa MO 2019, Problem 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Excursion in Mathematics by Bhaskaracharya 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"]If you read this, you will get to know some techniques to explore Diophantine Equations. Let's get an idea of m and n  by using the modulo technique. To get an idea of n, we must remove or eliminate m, to do that we take modulo 10. Observe that the equation demands to be taken modulo 10, given the numbers and it turns out that \( 19^n = (-1)^n = 1 mod 10 \). It implies that n must be even. Try to get an idea of m now. Also, (0,0) is a solution. So, we take both m and n as non-zero.

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To remove n, using the information that n is even, we can remove the variable n, taking modulo 4. So, \( 2m^2 + 1 = 1 mod 4  \) implies m must be even. Let m = 2p.

Observe that RHS is a square and LHS \( < 20^{2p} \).

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]The largest square \( < 20^{2p}\) is \( (20^{p} - 1)^2\). Thus,  \( LHS \leq (20^{p} - 1)^2 \). Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$,
which simplifies to   $p^2\ge20^{p-1}$. (*) [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Now, this turns out to be inequality and this results in the solution p = 1. This gives the only solution (2,2) as (m,n).[/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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