Prime Postulate

Question

Why is it the case that for every prime number $p_i$ there exist unique positive integers $m$ and $n$ such that $$ m\, p_i^2 + 1 = n \, p_{i + 1} $$ where $m\ne n$ and $m \le p_i$?

That is, why is it that there is some positive integer $m$, less than of equal to the $i$-th prime number $p_i$, such that when $p_i$ is multiplied by $m$, and one is added, the result equals the product of the next, $(i + 1)$-st prime number $p_{i + 1}$, and the positive integer $n$, for some $n\ne m$, and for each $i$ there is a unique way of achieving this?

Answer 1

It's false.

Here's a counterexample . . .

Let $p_i=43$.

Then the smallest positive integer $m$ for which there exists a positive integer $n$ satisfying $$mp_i^2 + 1=np_{i+1}$$ is $m=44$, which is greater than $p_i$.

The next two counterexamples are $$p_i = 173$$ $$p_i = 18869$$

Those are the only counterexamples with $p_i < 10^6$.

It's conceivable that there are no more counterexamples.

Note that for all $p_i$, the smallest qualifying positive integer $m$ is necessarily less than $p_{i+1}$, so there is a small window of opportunity for a counterexample, namely, $m$ needs to be greater than $p_i$, but less than $p_{i+1}$.

Answer 2

The Euclidean algorithm guarantees the existence of a unique $m\lt p_{i+1}$ for which $p_{i+1}\mid mp_i^2+1$. If you believe that each $m\lt p_{i+1}$ has an equal chance of being the unique such $m$ (which is obviously not true, if only because the properties of primes are deterministic, not random), then the probability that $m\le p_i$ is less than $1-{1\over p_{i+1}}$ (since there is at least one $m$ between $p_i$ and $p_{i+1}$, namely $p_{i+1}-1$). If you furthermore believe than these probabilities are independent from one prime to another (which isn't true either), then the probability that $m\le p_i$ from $i=N$ to $\infty$ is less than $\prod_{k=0}^\infty(1-{1\over p_{N+k}})=0$.

This is by no means a proof of anything, but it does suggest that the counterexamples found by quasi will continue to pop up. A more careful heuristic examination might even produce a formula for approximately how often.

Remark: Another reason not every $m\lt p_{i+1}$ has an equal chance is that we must have $\left(m\over p_{i+1}\right)=\left(-1\over p_{i+1}\right)$. But since $m=p_{i+1}-1$ satisfies this constraint, that's not a heuristic-killer.

Answer 3

Because

$mp_i^2 + 1 = np_{i+1} \iff$

$np_{i+1} - mp_i^2 = 1$

As $\gcd(p_i,p_{i+1}) = 1$

There always are such integers via Euclid's Algorithm.

If $m,n$ are such integers then $m' = m + k*p_{i+1}$ and $n' = n + kp_i^2$ for all $k$ are all such solutions. (Because $(m+kp_{i+1})p_i^2 +1= mp_i^2 +1+ kp_i^2p_{i+1} = np_{i+1} + kp_i^2p_{i+1} = (n + kp_i^2)p_{i+1}$.)

And $k$ so that $0 < m + k*p_{i+1} \le p_{i+1}$ is unique by division algorithm.

(You have a typo that $m \le p_i$. That need not be true.)

(If $n = m$ would imply $mp_i^2 + 1 = mp_{i+1} \implies p_i^2 +\frac 1m = p_{i+1}$ which means $m = 1$ so $p_i^2 + 1 = p_{i+1}$ which means there are no primes between $p_i$ and $p_i^2$ which is impossible.)