Number Theory (Thailand Math POSN 2nd round)

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Theorems allowed to use: - Divisibility, gcd, lcm, Prime numbers - Modular arithmetic - Chinese Remainder Theorem - Complete Residue System modulo n - Reduced Residue System modulo n - Euler's Theorem - Fermat's Little Theorem - Wilson's Theorem

1.) Prove that there are no positive integers \(n\) such that

\[2558^{n} \equiv 1 \pmod{2000^{n} - 1}\]

2.) Find the largest positive integer \(n\) such that for all positive integers\(a\),

\[a^{9} \equiv a \pmod{n}\]

3.) Find all solutions to linear congruence

\[2557x \equiv 2015 \pmod{1200}\]

4.) Let \(p\) be a prime number and \(N = \displaystyle \frac{10^{p} - 1}{9}\), prove that

\[N\times 10^{8p} + 2N\times 10^{7p} + 3N \times 10^{6p} + \dots + 9N \equiv 123456789 \pmod{p}\]

5.) Let \(p\) be a prime number, prove that the congruence

\[x^{2} \equiv -1 \pmod{p}\]

has a solutions if and only if \(p = 2\) or \(p \equiv 1 \pmod{4}\).

This note is a part of Thailand Math POSN 2nd round 2015

Note by Samuraiwarm Tsunayoshi
3 years, 8 months ago

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Solution to Q-4:

From Fermat's Little Theorem, we know that, for any integer \(a\) and a prime \(p\), we have \(a^p\equiv a\pmod{p}\). Then,

\[10^p\equiv 10\pmod{p}\implies 10^p-1\equiv 9\pmod{p}\implies N=\frac{10^p-1}{9}\equiv 1\pmod{p}~,p\neq 3\]

We have \(p\neq 3\) in that case since \(\gcd(9,3)=3\neq 1\) and \(3\) is the only prime that is not coprime to \(9\).

For \(p=3\), using divisibility rule of \(3\), we have \(N=\dfrac{10^3-1}{9}=111\equiv 0\pmod{3}\). Hence, we combine these two results as (with prime \(p\)),

\[N=\frac{10^p-1}{9}\equiv \begin{cases}1\pmod{p}~,~p\neq 3\\ 0\pmod{p}~,~p=3\end{cases}~\ldots~(i)\]

Trivial Observation 1: We note beforehand that \(123456789\equiv 0\pmod{3}\) using the divisibility rule of \(3\). As such, if we have \(x\equiv 0\pmod{3}\), we can also write that as \(x\equiv 123456789\pmod{3}\) for any integer \(x\).

Now, for positive integral \(n\), \(10^n\in\mathbb{Z}\) and hence, by FLT, we have \(10^{np}\equiv 10^n\pmod{p}\) for all primes \(p\) and positive integral \(n\). We use this result to form a modular congruence equation with the LHS of the statement to be proved modulo prime \(p\) as,

\[\textrm{LHS}\equiv N(10^8+2\times 10^7+3\times 10^6+\ldots +9)\equiv 123456789N\pmod{p}\]

Using \((i)\), this becomes,

\[\textrm{LHS}\equiv\begin{cases} 123456789\pmod{p}~,~p\neq 3\\ 0\pmod{p}~,~p=3\end{cases}\]

Using Trivial Observation 1, this becomes,

\[\textrm{LHS}\equiv 123456789\pmod{p}=\textrm{RHS}\]

\[\therefore\quad N\times 10^{8p} + 2N\times 10^{7p} + 3N \times 10^{6p} + \dots + 9N \equiv 123456789 \pmod{p}\]

Prasun Biswas - 3 years, 8 months ago

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Solution to Q no 3 : 2557x=2015(mod 1200) Since gcd (2557,1200)=1 there is an unique solution (mod 1200) By using the Euclidean Algorithm to write 1 as the linear combination of 2557 and 1200, such as 1200×228-2557(-107)=1 here 2557(-107)=1(mod 1200) so, 2557(-107×2015)=2015 (mod 1200) Since , -107×2015 = 395 (mod 1200) We conclude that x=395 ( modulo 1200) is an unique solution . so x =395

Abdullah Ahmed - 3 years, 2 months ago

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Nice solution. Also this question can be done by the CRT

Shrihari B - 2 years, 12 months ago

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prob 1 : first assume there is a positive number n such that 2558^n =1 (mod 2000^n -1) or, 2558^n-1=0( mod 2000^n-1) that means 2000^n-1 divides 2558^n-1 now , let p be the maximum prime divisor of 2000^n -1 so. 2000^n=1(mod p) let this p be also a maximum prime divisor of 2558^n-1 so. 2558^n=1 (mod p) then , 2000^n =2558^n or 2000= 2558 and it is not possible . So there are no integer n such that 2558^n=1(mod 2000^n-1)

Abdullah Ahmed - 3 years, 2 months ago

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In your solution, it seems to me that you assume that \(a^n\equiv b^n\pmod{p}\implies a=b\) where \(a,b,n\) are positive integers and \(p\) is a prime. But that doesn't necessarily hold true.

Consider \(a=3,b=4,p=11,n=10\). We have \(a\neq b\) but \(a^n\equiv b^n\pmod{p}\).

Prasun Biswas - 3 years, 2 months ago

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