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1 vote
1 answer
40 views

Why $xy=N\Rightarrow (((x−1)p−1\text{ mod }N)−1)\text{ mod} X=0$

Experimentally, I found that for $N=xy;\ 3<x<y$ where $x,y$ are prime numbers and for prime numbers, $p>x$ the below expression is always true: $$\Big(\big((x-1)^{p-1}\text{ mod }N\big) -1\...
Ilya Gazman's user avatar
  • 1,450
1 vote
4 answers
98 views

When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer

$a,b,m,x$ are positive integers. For which $x>0$ is $f(x)$ an integer? $$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$ I been trying to play with it, I changed it to: $$\frac{b^2m-a\left(b+x\right)}{a+m\...
Ilya Gazman's user avatar
  • 1,450
1 vote
1 answer
73 views

When $f(x) = \frac{ax + b}{a -x + 1}$ is an integer

Given that $a$ and $b$ are positive integers. and $$f(x) = \frac{ax + b}{a -x + 1}$$ is an integer, what integer values can x have? If I could only somehow move $x$ from numerator to the denominator ...
Ilya Gazman's user avatar
  • 1,450
1 vote
2 answers
81 views

Roots modulo a prime

Let us consider the following result: Note that in the ring of polynomials with coefficients $mod p$, the product $x(x-1)(x-2)...(x-p+1)$ has all possible roots, and equals $x^p-x$ by Fermat's Little ...
Safwane's user avatar
  • 3,854
1 vote
1 answer
33 views

Why is it that the degree of a subextension of $K(a^{1/p})/K$ must have a degree dividing $p$?

$p$ is a prime $K$ is of characteristic not $p$ $a∈K$ $a^{1/p}∉K$ $a$ is not a root of unity "Then if we pick an element $b∈K(a^{1/p})$, $b∉K$, then $K(b)/K$ is a non-trivial subextension, thus of ...
Sam's user avatar
  • 5,166
0 votes
2 answers
165 views

If $P,Q,R$ are natural numbers where $P$ and $Q$ are primes, and $Q$ divides $PR$, which of the following is true? [closed]

If $P,Q,R$ are natural numbers where $P$ and $Q$ are primes, and $Q$ divides $PR$, which of the following is true? $P\mid\,Q$ $P\mid\,R$ $P\mid\,QR$ $P\mid\,PQ$ I know that the last option is ...
Sriram D's user avatar
-1 votes
1 answer
131 views

Find largest small prime factor [closed]

Let $\mathcal{Q}=19^5+660^5+1316^5$ . We know that $25$ is a factor of $\mathcal{Q}$ . Find with a proof the largest prime factor of $\mathcal{Q}$ not exceeding of $10,000$. I found by computer ...
John's user avatar
  • 1,056
1 vote
3 answers
107 views

Manipulation of Large Numbers

I encountered a puzzling problem (though I don't remember where) to find the prime factorization of the number $$7^{100}-1$$ I think there may be some kind of trick or technique that one could use to ...
Franklin Pezzuti Dyer's user avatar
0 votes
1 answer
72 views

P-adic order of a prime

Let $p$ be a prime number and $x \in \mathbb{Z}_{0}$. We define the $p$-adic order of $x$ (notation: $ord_{p}(x))$ as the greatest exponent $\alpha \in \mathbb{N}$ so that $p^{\alpha}|x$. Now show ...
simp's user avatar
  • 893
2 votes
2 answers
821 views

How many Positive Integers divide $72^{100}$?

So I began by finding the prime factorization of $72$, which is: $$ 72 = 2^{3} \times 3^{2}$$ then exponentiating by $100$: $$ 72^{100} = 2^{300} \times 3^{200}$$ and so now the question is asking to ...
Numerical Disintegration's user avatar
3 votes
1 answer
168 views

Divisibility Problem: How many switches remain in their initial position?

I have been working on this problem: There is a set of $1000$ switches, which are ordered in a row so that each switch is given a distinct rank from $1$ to $1000$. For example, the $i$-th switch ...
alxrf's user avatar
  • 143
0 votes
2 answers
398 views

Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$

Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$. I am stuck with this proof. I thought at first that I need to use contradiction but that did not work. What do you ...
Jamie John's user avatar
3 votes
1 answer
64 views

About the divisors of totient numbers

Are there infinitely many integers that do not divide any totient number? My try: If $a|b$ then $\phi(a)|\phi(b)$, so the main question would be equivalent to asking wether there are infinitely many ...
user3141592's user avatar
  • 1,919
3 votes
2 answers
765 views

How many positive integers are divisors of at least one of $24^5, 20^6$, and $45^7$?

I have a vague idea of prime factorizing 24, 20, and 45, then distributing the exponent, find how many numbers are divisible by them (factors), and then subtract the overcounts. How do I do this?
Gerard L.'s user avatar
  • 2,561
1 vote
3 answers
78 views

If $a^5|b^3$, does $a^2|b$?

I have proven that $a^5|b^3$ implies that $a\mid b$ using prime factorization and showing that every prime factor of $a$ will divide every prime factor of $b$. I have to prove or disprove that the ...
mmm's user avatar
  • 1,849

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