## Some basic results in number theory — part 1

Theorem (1): The product of any n consecutive integers is divisible by $n!$.

Proof (1):

For $\frac {(m+1)(m+2) \ldots (m+n)}{n!}=\frac{(m+n)!}{m!n!}$, and to show that the last expression is an integer it is sufficient to show that any prime p which occurs in $m!n!$ to at least as high as a power in

$(m+n)!$. Thus, we have to show that

$I[\frac{(m+n)}{p}]+I[\frac{(m+n)}{p^{2}}]+I[\frac{(m+n)}{p^{3}}]+\ldots$ is greater than or equal to

$I[m/p]+I[m/p^{2}]+I[m/p^{3}]+\ldots +I[n/p]+I[n/p^{2}]+I[n/p^{3}]+\ldots$.

Note: The  Symbol $I[x/y]$: If  a is a fraction or an irrational number, the symbol $I(a)$ will be used to denote the integral part of a.

Now, $I[\frac{(m+n)}{p}] \geq I[m/p]+I[n/p]$, and the same is true if we replace p by $p^{2}$, $p^{3}$ in succession, hence the result in question.

Theorem (2): If n is a prime, then ${n \choose r}$ is divisible by n.

For by the preceding $n(n-1)(n-2)\ldots (n-r+1)$ is divisible by $r!$ and since n is a prime and r is supposed to be less than n, $r!$ is prime to n. Hence,

$r!$ is a divisor of $(n-1)(n-2)\ldots (n-r+1)$

and $n(n-1)\ldots (n-r+1)/r!$ is divisible by n.

Thus, if a is a prime, all the coefficients in the expansion of $(1+x)^{n}$ except the first and last are divisible by n.

Homework:

1) Find the highest power of 5  contained in $158!$

2) If n is an odd prime, the integral part of $(\sqrt{5}+2)^{n}-2^{n+1}$ is divisible by $20n$.

More later,

Nalin Pithwa

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