recategorized by
1,182 views
5 votes
5 votes

Let $m$ and $n$ be any two positive integers. Then, which of the following is FALSE?

  1. $m + 1$ divides $m^{2n} − 1$.
  2. For any prime $p$, $m^{p} \equiv  m (\mod p)$.
  3. If one of $m$, $n$ is prime, then there are integers $x, y$ such that $mx + ny = 1$.
  4. If $m < n$, then $m!$ divides $n(n − 1)(n − 2) \ldots (n − m + 1)$.
  5. If $2^{n} − 1$ is prime, then $n$ is prime.
recategorized by

1 Answer

Best answer
3 votes
3 votes

(a) $m$ divides $m^{2n}.$  Now we can observe that now if we divide $m^{2n}$ by $m+1$ then remainder will always be $1.$ Now to make remainder $0$ we have to subtract $1$ from $m^{2n}$ so in that way we can say that $m+1$ divides $m^{2n}-1.$

(b) It is popular theorem by Mathematician Fermat

https://en.wikipedia.org/wiki/Fermat%27s_little_theorem

(c) To make $mx+ny=1,$ $n$ and $m$ both must have to be relatively prime. If only one is prime and then other, when it is a multiple of that prime, sum cannot be 1.

like $5x+ny=1$ if $n$ is $10,15,20\ldots$ result cannot be $1.$ Basically the theorem is

For any positive integers $a$ and $b$, there exist integers $x$ and $y$ such that
$mx+ny= gcd(m, n).$

Now when $m$ and $n$ are relatively prime then their $\gcd$ will be $1$ so eventually equation would be like $mx+ny=1$

(d) We know that ${}^nC_m$ means choose $m$ out of $n$ and result of this will always be integer. If we see carefully what's being asked is ${}^nC_m,$ which is always an integer so yes $m!$ will have to divide $n\times (n-1)(n-2)\ldots (n-m+1)$.

(e) https://en.wikipedia.org/wiki/Mersenne_prime

So, option C must be false.

edited by
Answer:

Related questions

3 votes
3 votes
1 answer
4