I’ve been reviewing the following: $$v_i = \left\lfloor\frac{ip_k\#}{p_{k+1}}\right\rfloor + c_i$$ where: $c_i \in \left\{1,2\right\}$ so that $v_i$ is odd and $v_ip_{k+1} > ip_k\# > (v_i-c)p_{k+1}$ $i$ is an integer such that $1 \le i \le p_{k+1}-1$ $\left\lfloor\frac{a}{b}\right\rfloor$ is a floor function I hit a result that surprises me. Let $[v_i]$ be a residue modulo $p_k\#$ […]

I don’t know how to determine it… any hints?

What does $p^\alpha\| n$ mean ? I saw this in Euler totient function, $$\varphi(n)=\prod_{p^\alpha\| n}p^\alpha(p-1).$$

If $a$ and $b$ are positive integers such that $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$. By contradiction assuming that $a <b$ then it follows that $2 (a +1) <A <2 (b +1)$. This means that $A = 2a + r,$ $r$ taking values $3, 4, …, 2b-2a +1$. From the equality […]

Let $\sigma(x)$ denote the sum of the divisors of $x$, and let $I(x) = \sigma(x)/x$ be the abundancy index of $x$. A number $y$ is said to be almost perfect if $\sigma(y) = 2y – 1$. In a preprint titled A Criterion For Almost Perfect Numbers Using The Abundancy Index, Dagal and Dris show that […]

I need to make the proof for this 1:$$1^2 + 2^2 + 3^2 + … + n^2=\frac{(n(n+1)(2n+1))}{6}$$ By mathematical induction I know that, If P(n) is true for $n>3^2$ then P(k) is also true for k=N and also P(k+1) must be true. Then, 2:$$1^2 + 2^2 + 3^2 + … + k^2=\frac{(k(k+1)(2k+1))}{6}$$ Substitution: 3:$$1^2 + […]

The two forms are: $\ 3x^2 + (6y-3)x – y\ $ $\ 3x^2 + (6y-3)x + y – 1, \ \ x,y \in \mathbb{Z}^{+}$ For example: $\ \ \ 5 = \ 3*1^2 + (6*1-3)*1 – 1\ $ ,when $\ x = y = 1\ $,of the two forms $\ 13 = \ 3*1^2 + […]

I recently came across this relation valid for all positive integers …which is $n!\leq n^n$ while it’s proof is very easy by basic induction …but I wanted to know for what values of n .. the equality($=$) holds..? I tried putting some values but I found only n=1 is satisfying it…are there any other values […]

I’m trying to show that if $\operatorname{gcd}(a,b) = 1$, then $\operatorname{gcd}(ab,a+b)=1$. I’ve tried to use the gcd properties: $$\operatorname{gcd}(a,b)=1 \implies \operatorname{gcd}(a,a+b)=1\\\operatorname{gcd}(a,b)\implies\operatorname{gcd}(ab,b^2)=b$$ but I got stuck. Any hint will help.

I’ve checked a lot of the congruency posts and haven’t seen this one yet, so I’m going to ask it. If there is a related one, I’d be happy to see it. Let $x \equiv r\pmod{m}, x \equiv s\pmod{(m+1)}$. Prove $$x \equiv r(m+1)-sm \pmod{m(m+1)}$$ So with the given conditions, we know $$x=mk_1+r$$ Then $mk_1+r \equiv […]

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