# Tag Info

11

$$n(n+1)(n+2)(n+3)\equiv (-k-3)(-k-2)(-k-1)(-k)$$ $$\equiv (-1)^4(k+3)(k+2)(k+1)k\equiv k(k+1)(k+2)(k+3)\pmod{n+k+3}$$ It's true for all integers $n,k$ with no restrictions.

6

Since $$n+3=n+k+3-k\qquad \text{and}\qquad k+3=n+k+3-n$$ we have \begin{align} n(n+1)(n+2)(n+3)&=n(n+1)(n+2)(n+k+3)-n(n+1)(n+2)k\\ k(k+1)(k+2)(k+3)&=k(k+1)(k+2)(n+k+3)-k(k+1)(k+2)n \end{align} Then it will be sufficient to show that $n+k+3$ divides $k(k+1)(k+2)n-n(n+1)(n+2)k$ But \begin{align} k(k+1)(k+2)n-n(n+1)(n+2)k&=nk(k^2+3k-n^2-3n)\\ ...

3

If $P$ is a polynomial with integer coefficients, then $$a-b \mid P(a)-P(b)$$ for integers $a$ and $b$. The desired conclusion follows by applying this with $a=n$, $b=-k-3$ and $$P(x) = x(x+1)(x+2)(x+3).$$ Indeed, we have $P(n)=n(n+1)(n+2)(n+3)$ and $P(-k-3) = (-k-3)(-k-2)(-k-1)(-k) = k(k+1)(k+2)(k+3)$.

3

Use Euclidean algorithm. $$\frac{a^m-1}{a-1}=a^{m-1}+a^{m-2}+\cdots + 1$$ $$\equiv 1^{m-1}+1^{m-2}+\cdots + 1\equiv m\pmod{a-1}$$ Therefore $\gcd\left(\frac{a^m-1}{a-1},a-1\right)=\gcd(m,a-1)$. More generally: $$\gcd\left(\frac{a^m-b^m}{a-b},a-b\right)=\gcd\left(m(\gcd(a,b))^{m-1},a-b\right)$$

2

This is true for arbitrary integers $p$ and $q$, not just distinct odd primes. It suffices to show that every prime power dividing both $\mathrm{lcm}(p-1,q-1)$ and $pq-1$ must also divide $p-1$ and $q-1$. Suppose $\ell^n$ is a prime power dividing $\mathrm{lcm}(p-1,q-1)$ and $pq-1$. The first divisibility implies $p\equiv 1$ or $q\equiv 1\mod\ell^n$. The ...

2

The proper term to use is "removable discontinuity". You're correct that the expression equals 1 in pretty much every case - still, 0 is not a part of that function's domain. The function you gave is equivalent to this: $$f(x)=1, x\neq0$$

2

$$\overline{abcba}\equiv 10^4\cdot a+10^3\cdot b+10^2\cdot c+10^1\cdot b+a$$ $$\equiv (-1)^4a+(-1)^3b+(-1)^2c+(-1)^1b+a$$ $$\equiv a-b+c-b+a\equiv 2a-2b+c\equiv 0\pmod{11}$$ To minimize $\overline{abcba}$, let $a=1$ and $b=0$. Then $c\equiv 9\pmod{11}$, so $c=9$. And in fact $\overline{10901}$ works.

2

Let's say that $a_n$ represents the number of numbers with n digits, which are divisible by 5 and have no 2 consecutive digits identical. Now, consider the case ($n>2$) where the 2nd digit is nonzero and equal to $p$. Thus, the first digit can anything except $p$ (so as to have non-identical consecutive digits) and $0$ (so that it remains a $n$ digit ...

1

The quadratic residues mod $3$ are $0$ and $1$. The only sum of two of these that is $0$ mod $3$ is $0+0$. Therefore, $a^2\equiv0\pmod3\implies a\equiv0\pmod3$ and $b^2\equiv0\pmod3\implies b\equiv0\pmod3$.

1

Question : Is it true that for every pair of primes $p<q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$ ? Answer. yes. I will not provide a proof for this theorem here, but I will point to some references for the proof. Actually a generalization of your claim is true: for ever pair of integers $a<b$ there are only ...

1

First Proof Claim. $kd\mid d$. Proof $d\mid a \land k\mid r \implies kd\mid ar$ $d\mid b \land k\mid s \implies kd\mid bs$ $\therefore kd\mid ar+bs\implies kd\mid d\implies ??$ Second Proof But I think that the proof can be done in much simple way if you just notice that $r\left(\dfrac{a}{d}\right)+s\left(\dfrac{b}{d}\right)=1$. ...

1

Let $l:=\text{lcm}(p-1,q-1)$ and $g:=\gcd(p-1,q-1).$ Since $lg=(p-1)(q-1)=(pq-1)+(p+q-2)$ and since $k\mid lg,\;pq-1$ then $k\mid p+q-2=(p-1)+(q-1).$ If $t$ is any prime factor of $k$ then $t\mid lg=(p-1)(q-1)$ and hence either $t\mid p-1$ or $t\mid q-1.$ In the first case we have $t\mid q-1$ because $k\mid(p-1)+(q-1)$ and the same happens in the second ...

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