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# 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
1 year, 10 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,

$\textrm{LHS}\equiv 123456789\pmod{p}=\textrm{RHS}$

$\therefore\quad N\times 10^{8p} + 2N\times 10^{7p} + 3N \times 10^{6p} + \dots + 9N \equiv 123456789 \pmod{p}$ · 1 year, 9 months ago

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 · 1 year, 3 months ago

Nice solution. Also this question can be done by the CRT · 1 year, 1 month ago

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) · 1 year, 3 months ago

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 $$a^n\equiv b^n\pmod{p}$$. · 1 year, 3 months ago