I am trying to prove the following: $p + 1$ consecutive integers contain at least two invertible elements modulo $m = 3 \cdot 5 \cdots ( p – 2 ) \cdot p$, where $p$ is odd. I shared my idea in an answer to my own question..

I am having problems solving this question : When n is divide by 4 the remainder is 2 what will the remainder be when 6n is divided by 4 ? Ans=$0$ Here is what I have got so far $\frac{n}{4} => Remainder ~ 2$ so we get $n=4q+2$ $\frac{6n}{4} => Remainder ~ ?$ so we […]

There have been a number of posts about the harmonic series, e.g. not being an integer for any number of terms. Edit: Below I try to prove that not only H(n) but also H(2n)-H(n) is not an integer. I note e.g. that H(2n)-H(n) must contain a term 1/b with b prime according to Bertrand´s theorem […]

I am utterly new to modular arithmetic and I am having trouble with this proof. $$2222^{5555}+5555^{2222}=3333^{5555}+4444^{2222} \pmod 7$$ It’s because $2+5=3+4=7$, but it’s not so clear for me with the presence of powers. Maybe some explanation would help. EDITED Some serious typo EDIT Since some arguments against it appear here is : WolframAlpha EDIT Above […]

Let $q$ be an odd composite integer and $\sigma(q)$ the sum of the positive divisors of $q$. For what $q$ is it true that $$(\sigma(q)-q) \mid (q-1) \;?$$ If $q$ is prime, it is clear that it is true. Is it only true for all primes $q$? Thanks for the help.

If $F_n$ denotes the $n$-th Fibonacci number ($F_0 = 0, F_1 = 1, F_{n+2} = F_{n+1} + F_n$), show that $5^n$ divides $F_{5^n}$.

Show that if $c\mid a-b$ and $c\mid a’-b’$ then $c\mid aa’-bb’$: I have tried to express $aa’-bb’$ in terms of $a-b$ and $a’-b’$ as: $2(aa’-bb’) = (a-b)(a’+b’)+(a’-b’)(a+b)$ which implies that $(aa-bb’) = (a-b)(a’+b’)/2+(a’-b’)(a+b)/2$. However, there is no guarantee that all these terms are divisible by 2; how can I proceed?

if $p$ is odd prime and $\{a_1,…,a_p\},\{b_1,…,b_p\}$ are complete residue system modulo $p$ how to prove $\{a_1b_1,…,a_pb_p\}$ is not complete residue system modulo $p$. complete residue system modulo $p$

Let $N_n$ be an integer formed of $n$ consecutive $1$s. For example $N_3 = 111,$ $N_7 = 1 111 111.$ Show that $N_n \mid N_m$ if and only if $n \mid m.$

I just stumbled across the identity $$1\times2+2\times3+\cdots+34\times35=119\times120,$$ which made me wonder about integral solutions to $$\sum_{k=1}^mk\times(k+1)=n\times(n+1).$$ The left hand side can be rewritten as $$\sum_{k=1}^mk^2+\sum_{k=1}^mk=\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}{2}=\frac{m(m+1)(m+2)}{3},$$ which shows that the original problem is equivalent to solving $$m(m+1)(m+2)=3n(n+1),$$ over the integers. This looks nicer already, but I’m still unable to determine its integral solutions. The latter equation […]

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