Difference between revisions of "2024 INMO"
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∼Lakshya Pamecha | ∼Lakshya Pamecha | ||
+ | ==Problem 3== | ||
+ | Let p be an odd prime number and a,b,c be integers so that the integers <cmath>a^{2023}+b^{2023}, b^{2024}+c^{2024}, c^{2025}+a^{2025}</cmath> are all divisible by p. Prove that p divides each of <math>a,b,c</math>. | ||
+ | ==Solution== | ||
+ | If <math>p\vert{a}</math> \Rightarrow <math>p\vert a^{2023}</math> and <math> p \vert a^{2023}+b^{2023}</math> \Rightarrow p\vert b \Rightarrow <math>p\vert b^{2024}</math> and <math>p\vert b^{2024}+c^{2024}</math> \Rightarrow <math>p\vert c</math>.\\ | ||
+ | Therefore, if <math>p</math> divides one of <math>a,b,c</math> it will divide all of them.\\ | ||
+ | Assume that <math>p</math> does not divide <math>a, b </math>or <math>c</math> | ||
+ | Set | ||
+ | <math></math> a^{2023} &\equiv k \pmod{p} \Rightarrow b^{2023} \equiv -k \pmod{p} \\ b^{2024} &\equiv -bk \pmod{p} \Rightarrow c^{2024} \equiv kb \pmod{p}\\ c^{2025} &\equiv kbc \pmod{p}\Rightarrow a^{2025} \equiv -kbc \pmod{p}\\$<math> | ||
+ | <cmath>\Rightarrow \boxed{a^2 &\equiv -bc \pmod{p}}</cmath> | ||
+ | Now we see that | ||
+ | </math><math>(a^{2023})^2 &\equiv (b^{2023})^2 \pmod{p}\\ (-bc)^{2023} &\equiv (b^2)^{2023} \pmod{p}\\ \Rightarrow -c^{2023} &\equiv b^{2023} \pmod{p}\; \text{and} \; b^{4048} \equiv c^{4048}\pmod{p}\$</math> | ||
+ | \text{So}, | ||
+ | <cmath>\boxed{b^2 &\equiv c^2 \pmod{p}}</cmath> | ||
+ | This gives us to 2 cases:\\ | ||
+ | Case I: | ||
+ | <cmath>b-c \equiv 0 \pmod{p} | ||
+ | \Rightarrow b^{2024} \equiv c^{2024} \pmod{p} | ||
+ | \Rightarrow 2b^{2024} &\equiv 0 \pmod{p} \Rightarrow p\vert b </cmath> | ||
+ | Case II: | ||
+ | <cmath>b+c \equiv 0\pmod{p} \Rightarrow -bc \equiv c^2 \pmod{p} \Rightarrow a^2 \equiv c^2 \pmod{p} \\\Rightarrow a \equiv c \pmod{p} \;\;\text{OR} \;\; a \equiv -c \pmod{p}</cmath> | ||
+ | On checking for both cases we get <math>p\vert{a}</math> which implies <math>p\vert{b}</math> and <math>p\vert{c}</math>. |
Revision as of 13:27, 25 April 2024
==Problem 1
\text {In} triangle ABC with , \text{point E lies on the circumcircle of} \text{triangle ABC such that} . \text{The line through E parallel to CB intersect CA in F} \text{and AB in G}.\text{Prove that}\\ \text{the centre of the circumcircle of} triangle EGB \text{lies on the circumcircle of triangle ECF.}
Solution
https://i.imgur.com/ivcAShL.png To Prove: Points E, F, P, C are concyclic
Observe: Notice that because . Here F is the circumcentre of because lies on the Perpendicular bisector of AG is the midpoint of is the perpendicular bisector of . This gives And because Points E, F, P, C are concyclic. Hence proven that the centre of the circumcircle of lies on the circumcircle of .
∼Lakshya Pamecha
Problem 3
Let p be an odd prime number and a,b,c be integers so that the integers are all divisible by p. Prove that p divides each of .
Solution
If \Rightarrow and \Rightarrow p\vert b \Rightarrow and \Rightarrow .\\ Therefore, if divides one of it will divide all of them.\\ Assume that does not divide or Set $$ (Error compiling LaTeX. Unknown error_msg) a^{2023} &\equiv k \pmod{p} \Rightarrow b^{2023} \equiv -k \pmod{p} \\ b^{2024} &\equiv -bk \pmod{p} \Rightarrow c^{2024} \equiv kb \pmod{p}\\ c^{2025} &\equiv kbc \pmod{p}\Rightarrow a^{2025} \equiv -kbc \pmod{p}\$$<cmath>\Rightarrow \boxed{a^2 &\equiv -bc \pmod{p}}</cmath> Now we see that$ (Error compiling LaTeX. Unknown error_msg)$(a^{2023})^2 &\equiv (b^{2023})^2 \pmod{p}\\ (-bc)^{2023} &\equiv (b^2)^{2023} \pmod{p}\\ \Rightarrow -c^{2023} &\equiv b^{2023} \pmod{p}\; \text{and} \; b^{4048} \equiv c^{4048}\pmod{p}$ (Error compiling LaTeX. Unknown error_msg) \text{So},
\[\boxed{b^2 &\equiv c^2 \pmod{p}}\] (Error compiling LaTeX. Unknown error_msg)
This gives us to 2 cases:\\ Case I:
\[b-c \equiv 0 \pmod{p} \Rightarrow b^{2024} \equiv c^{2024} \pmod{p} \Rightarrow 2b^{2024} &\equiv 0 \pmod{p} \Rightarrow p\vert b\] (Error compiling LaTeX. Unknown error_msg)
Case II: On checking for both cases we get which implies and .