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There are two parts I am having trouble getting started.

A. Prove that $n_1, n_2,...,n_k\in\mathbb{N}$ are each at least $2$ then $n=n_1n_2...n_k+1$ is not divisible by any numbers $n_1, n_2,...,n_k$.

B. Prove that the truth of the negation leads to a contradiction. (Use theorem: For all $a,b\in\mathbb{N}$ there exist a unique quotient $q$ and remainder $r$ in $\mathbb{Z^+}$ such that we have both $a=qb+r$ and $0\leq r<q$.)

For part A, I started with, given $k\in\mathbb{N}$ and $n_1, n_2,...,n_k\geq1$, I'll show that $\forall i$, $n_i \nmid n=n_1n_2...n_k+1$ to set it up, but I'm not sure how to actually go about starting it.

For part B, I know that the negation is $\exists n\in\mathbb{N}$ s.t. $\forall m\in\mathbb{N}$ either $m\leq n$ or $m$ is not prime, but again I'm not sure what I should do to start the proof or exactly how to incorporate that theorem.

rtybase
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A.A.
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3 Answers3

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This is a proof by contradiction: assume there is a largest prime

For part (A), we construct $N=n_1n_2n_3...n_k+1$ for $n_1,n_2,n_3,...\in\mathbb{N}$ and n_k the largest prime number.

From there, we need only one fact: if $q\in\mathbb{N}$, and if $q|N-1$, then $q\nmid N$. This is easily proven with modular arithmetic, but one can see it logically (if q|N-1, then the next number it will divide is the next multiple of $q$, or $N-1+q>N$ for $q>1$). Alternatively stated with modular arithmetic,

$q|N-1\Longleftrightarrow N-1\equiv 0\mod q\Longleftrightarrow N\equiv 1\not\equiv 0\mod q\Longrightarrow q\nmid N$.

For part (B), we note that $n_1|N-1$, $n_2|N-1$, $n_3|N-1$, ... From the above proven fact, $n_1\nmid N$, $n_2\nmid N$, $n_3\nmid N$, ..., and $n_k\nmid N$. But because $n_k$ is the largest prime, and all numbers below $n_k$ also do not divide $N$ no know prime number divides N.

However, it is a basic fact that all integers must have at least one prime factor. Therefore, either $N$ is prime or has prime factors which are all larger than the $n_k$.Therefore, we have a contradiction; there is no largest prime $n_k$.

This is equivalent to the statement you're trying to prove.

Tejas Rao
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Suppose that there is a highest prime, say $n$ and now consider $a_n=n!+1$. Since every integer greater than 2 has a prime divisor then $a_n$ has a prime divisor $p, p\leq n$. Since $p$ divise $n!$ so: $p|a_n-n! \Rightarrow p|1$ contradiction. Therefore there is no highest prime number and $\forall, m$ there is a prime $p> m$.

DINEDINE
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Regarding Part(A): The question in the title and the Part(A) question are not the same. Other responses seem to have assumed that the $n_k$ are consecutive, or perhaps are prime, which are not conditions of the question. For example, for $n_1=8,\ n_2=10$, $\ n_1n_2+1=81$ whose only prime factor is $3$, which is smaller than either $n_k$. It is nonetheless true in this case that $n_k\not \mid 81$.

So to answer Part(A), we need to recall that each $n_i=\prod p_j$ and $p_j\mid n_i \Rightarrow p_j\mid (n_1n_2\dots n_k)$. But $p_i\mid M \Rightarrow p_i\not \mid (M+1)$, because $p_i\not \mid 1$. So in this instance $p_i\not \mid (n_1n_2\dots n_k+1)$.

By the fundamental theorem of arithmetic, we know that $(n_1n_2\dots n_k+1)=\prod q_m$, and we can see that $p_j\ne q_m$. Since no prime factor of any $n$ divides $(n_1n_2\dots n_k+1)$, no $n$ can divide that number. But without further information, we can say nothing about the magnitudes of $q_m$ with respect to any $n$.

Keith Backman
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