A Beautiful Equation using Positive Integers from Singapore Math Olympiad 2008

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Understand the problem

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Find all pairs of positive integers $ (n,k)$ so that $ (n+1)^k-1=n!$.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.7"][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"]Singapore MO 2008[/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"]Number Theory[/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"]Medium[/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"]An Excursion in Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" 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

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Note that $latex n+1$ has to be a prime, because any proper prime divisor of $latex n+1$ would divide $latex n!$ too.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Say $latex n+1=p$. Note that, for large enough $latex p$, both $latex \frac{p-1}{2}$ and $latex 2$ are factors of $latex (p-1)!$. Thus, the factor $latex (p-1)$ occurs twice in the expansion of $latex (p-1)!$.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Prove that, if $latex (p-1)^2|p^k-1$ then $latex p-1|k$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]Hint 3 implies that $latex p-1\le k$. Hence $latex p^{p-1}-1\le p^k-1=(p-1)!$. This is obviously false. Hence, $latex p$ has to be small enough to avoid this situation. It is avoided precisely when $latex p\in\{2,3,5\}$. This corresponds to $latex n\in\{1,2,4\}$. The equations to be solved are $latex 2^k=2, 3^k=3$ and $latex 5^k=25$. Hence, the solutions are $latex \{(1,1),(2,1),(4,2)\}$.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" 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)

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Similar Problems

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