Can you hint me on how to show that $2(p-3)!\equiv -1\pmod{p}$, for $p>2$ prime. I that Wilson’s theorem says that $(p-1)!\equiv-1\pmod{p}$, and that $(p-3)!=(p-3)(p-2)(p-1)!$, but I’m not seeing how to fit this together.

How to prove following statement : Conjecture: An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$ if and only if $n$ is a prime number . If $x-y=m$ , where $m>1$ then $m \mid n$ Proof : $n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$ $\Rightarrow n=m\cdot (2\cdot y+m) […]

A recent question brought up the phenomenon that for some nonnegative integers $n$, $$ R(n^2) = R(n)^2, $$ where $R(n)$ means the digit reversal of $n$ (sometimes written $\overline{n}$). Some care is needed with regards to how to handle leading or tailing zeroes. However you decide the issue, one can see that it does not […]

To find all solutions greater than or equal to $1$ of a linear equation in the form $$x_1+x_2+x_3+\cdots+x_k=n ,$$ the number of them is $\binom{n-1}{k-1}$. If I need all solutions to be greater or equal to $0$ why is it then $\binom{n+k-1}{k-1}$? Please help. Thanks.

Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$. I know that for proving the $\gcd(a,b) = c$ you need to prove $c|a$ and $c|b$ $c$ is the greatest number that divides $a$ and $b$ Number 2 is what I’m struggling with. Does anybody have any ideas?

For what integers $n$ does $\phi(2n) = \phi(n)$? Could anyone help me start this problem off? I’m new to elementary number theory and such, and I can’t really get a grasp of the totient function. I know that $$\phi(n) = n\left(1-\frac1{p_1}\right)\left(1-\frac1{p_2}\right)\cdots\left(1-\dfrac1{p_k}\right)$$ but I don’t know how to apply this to the problem. I also know […]

Suppose $p$ and $q$ are prime numbers, $n>1$ and $m>1$ are positive integers. Solve the following Diophantine equation:$$(p+q)^n-p^n-q^n+1=m^{p-q}$$I made this problem and I was trying to find $p$ and $q$ first. Looking mod $p$, $q$ and $p+q$ gives$$m^{p-q} \equiv 1 \pmod {pq(p+q)}$$Does this provide useful information about $p$ and $q$?

Find the Prime Factorization of $\varphi(11!)$ What I did: $\varphi(11!)=\varphi(11)\cdot\varphi(10)…\varphi(1)$ $$\varphi(11)=2\cdot 5\\ \varphi(10)=2^2\\ \varphi(9)=3\cdot 2\\ \varphi(8)=2^2\\ \varphi(7)=2\cdot 3\\ \varphi(6)=2\\ \varphi(5)=2^2\\ \varphi(4)=2\\ \varphi(3)=2\\ $$ $$\Longrightarrow\text{ the answer is } 2^{12}\cdot 5\cdot 3^2$$ $11!$ $\varphi(11!)=\varphi(39916800)=8294400$ $8294400=2^{12}3^45^2$ Where am I wrong? Is there a simpler way to solve this?

I know the method of continued fractions to solve the Pell’s equation. I need help turning $2b(b-1) = t(t-1)$, with $b, t$ as integers, into the form $x^2 – ny^2 = 1$, if possible. My attempt is $(t-1/2)^2 – 2(b – 1/2)^2 = -1/4$, but those aren’t integer solutions anymore.

I want to prove: $$\text{If }a^2|z^2,\text{ then }a|z.$$ Assume everyone is a positive integer, etc. Unless I’m deluding myself, this is pretty easy to show using unique prime factorization. But I want to do it without using primes or the (usual statement of the) FTA. That is, using coprime is fine, using the so-called Bezout […]

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