Dtjytyk

Can you provide a counterexample for a claim given below?

Inspired by Puzzle 937 I have formulated the following claim:

For any $n > 0$ let $B = p_1 cdot p_2 cdot .... cdot p_n$ be the product of the first $n$ primes. Let $X$ be the smallest number, bigger than $B^k/p_{n+1}$ and coprime to $B^k$, where $k$ is a fixed positive integer. Define the number $m_n$ as $X cdot p_{n+1}-B^k$ , then $m_n$ is either $1$ or prime.

Try it for yourself!

I was searching for counterexample using the following PARI/GP code:

CE(lb,ub,k)={

for(n=lb,ub,

B=prod(i=1,n,prime(i));

X=ceil((B^k)/prime(n+1));

while(gcd(X,B^k)!=1,

X++);

m=X*prime(n+1)-B^k;

if(!(ispseudoprime(m) || m==1),print(m)))

}

I propose a generalization. Let $B = p_1^{k_1} cdots p_n^{k_n} $ where $k_i gt 0$ are fixed, except let $a = p_{n+1}$ be the same power (namely $1$) as in the original problem.

The set $S = { x in Bbb{Z} : (x, B) = 1}$ forms a saturated mult. closed subset of the ring $Bbb{Z}$.

This is true since if $1 in S$ and if $x, y in S$ then their product is also such that $(xy, B) = 1$ and if $xy in S$ then clearly $(x, B) = 1 = (y, B)$ since taking factors introduces nothing new.

If you minimize $x gt dfrac{B}{a}f, x in S$ then you equivalently minimize $ax gt B, ax in aS$ and since $k$ is fixed that is the same as minimizing $ax +$ any constant or $ax - B^k gt 0$ or ask what is the minimum positive element of $aS - B$?

If $n = ay - B$ is any element then if $n in aS$ we have $n = az$ or $a(y-z) = B$ or $a mid B$ a contradiction, thus $n in Bbb{Z} setminus aS$ and we have $aS - Bsubset Bbb{Z} setminus aS$.

Note that $aS = (a) cap S$.

True since clearly $aS subset (a)$ and since $a in S$ by definition ie. $(a, B) = 1$ we have $aS subset S$. Going the other direction: $x in S$ and $x = ay$ means that $y in S$ since $S$ is saturated! And so $x in aS$.

Thus we've proved that $n in (p_1) cup cdots cup (p_n) cup Bbb{Z}setminus(a) $ since by the link in the first sentence the complement of $S$ must be the union of the prime ideals making up $B$.

But $n notin (p_1) cup cdots (p_n)$ since other wise $ax = n + B in (p_i)$ for some $p_i$ contradicting $(B, x) = 1$. Thus $n in (Bbb{Z}setminus (a)) cap S = S'$ which is another saturated multiplicative set since $a$ is prime.

$S'$ is a saturated multiplicative set properly contained in $S$.

Thus you have a map $f(w) = aw - B : S to S'$ which is clearly injective and strictly increasing.

If you minimize the expression $f(x) gt 0, x in S$ then you've minimized $x$, and if you minimize $x in S$ such that $f(x) gt 0$ then $f(x)$ is a minimum in $f(S)$.

Proof. Exercise.

We have that $f(S) subset S'$ but we don't have the other direction. Let's show that $f(S)$ has sufficient properties for our purposes.

If $f(x)f(y) in f(S) subset S' subset S$, then if $f(x) notin f(S)$ then $x notin S$ so that $ax notin S$ by saturation, meaning $ax in Bbb{Z}setminus S = (p_1) cup cdots cup (p_n)$. Thus, since those primes make up the factors of $B$,
$ax = p_i z$ for some $i=1..n, z in Bbb{Z}$ and so $f(x) = ax - B in (p_i)$, which means that $p_i mid f(x)f(y) in f(S) subset S$ a multiplicative generated by strictly non-$B$ prime factors. This is a gross contradiction! Therefore

$f(S)$ is a "saturated subset" of $S'$.

The set $f(S)_{ gt 0} equiv { f(x) gt 0 : x in S}$ has the property that if $x, y gt 0$ and $xy in f(S)_{gt 0}$ then $x, y in f(S)_{gt 0}$ so its saturated in that sense.

If $1 in f(S)_{gt 0}$ then the minimum sought is indeed $1$ and we're done so assume that $1$ is not part of this saturated subset. Now if $xy in f(S)_{gt 0}$ and $x, y neq 1$ then we must have that $x, y$ are both in $f(S)_{gt 0}$ by saturation. This means that:

The minimal element of $f(S)_{gt 0}$ is indeed prime since if it were not, you could take a smaller element.

$square$

Corollary. Since $f(S) subset S' = Bbb{Z} setminus (p_1) cap cdots cap Bbb{Z}setminus (p_{n+1})$ we have that $min f(S)_{geq 0}$ is either $1$ or a prime $p_m gt p_{n+1}$.

Since the exponents give you "a lot of freedom", I conjecture that:
$$min ((aS - {p_1^{k_1} cdots p_{n}^{k_n} : k_i gt 0 })cap [2, infty)) = p_{n+2}$$, or the smallest prime attainable over any of the nonzero exponents is the next prime.