Number Theory, France IMO TST 2012, Problem 3

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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"]Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that:
\[a^p+b^p=p^c.\][/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" _builder_version="4.0" open="on"]France IMO TST 2012, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="4.0"]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"]Challenges and Thrills of Pre College Mathematics[/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"]Observe that  we will try to fundamental solutions to \( a^p + b^p = p^c\).  A fundamental solution (a,b,c,p) gives infinitely many solutions \( (a.p^k, b.p^k, c+k, p)\). A fundamental solution is, therefore (a,b,c,p) if gcd(a,b) = 1. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]We will focus on the fundamental solutions. We will deal with two cases: Case 1: p = 2 The equation reduces to \( a^2 + b^2 = p^2 \).  As gcd(a,b) = 1, it implies a and b are odd. Now any odd square = 1 mod 4. So, \( a^2 + b^2 = 2 mod 4 \). Hence, the only fundamental solution is (1,1,1) = (a,b,c)  We have that the following solutions are: $(1, 1, 1)$ and $\left(2^k, 2^k, 2^{2k+1}\right)$.  

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Case 2: p is an odd prime This now requires the idea of Lifting the Exponents. Please read here if you don't know it. It is an advanced technique to deal with Diophantine Equations. Let's check that the conditions of the LTE are satisfying here. p is an odd prime. gcd(a,b) = 1. p doesn't divide a or b as we are looking for fundamental solutions. \( a^p = a mod p; b^p = b mod p \). Hence, \( a^p + b^p = a + b mod p \). So, p | a+b, and p don't divide a or b.  Hence, we can apply LTE. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Using the LTE idea, we get  $1+v_p(a+b)=v_p\left(a^p+b^p\right) = c$ Then since $a+b | a^p+b^p$, we have that $a+b = p^{c-1}$, so $a^p+b^p = pa+pb$ Now, you see this can't happen for large p, as the LHS is exploding too fast like exponential as p increases and RHS is linear in p. So, we will apply inequality to prove this and find a bound for p for which it works and search in that bound. Note that $a^p \geq pa, b^p \geq pb$ or $a^{p-1} \geq p, b^{p-1} \geq p$ if $a, b \geq 2$. Therefore, we must have that $a=1, b=p^{c-1}-1$ (or vice versa). But clearly $\left(p^{c-1}-1\right)^p > p^c$, so no solution. 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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