Prime Postulate 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? 
 A: 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 $p_{i+1}$.
A: 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.
A: 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.)
