Improving Gaps in Primes Proof to Include Powers of Primes

Let $a!+b=p^{k}$ for some integers $1<b≤a$ and $p$ prime. Since $b≤a, b|a!$, and so $b|a!+b= p^{k}$. Thus, $b=p^{m}$ for some $m<k$. If $m>1$, then $p,p^{2},⋯,p^{m}<a$, and so $p^{m+(m-1)+⋯+1}|a!=p^{k}-p^{m}=p^{m}(p^{k-m}-1)$, and $p^{(m-1)+(m-2)+⋯+1}|p^{k-m}-1$. This implies that $p^{k-m} \equiv 1 \pmod{p}$. But this only works if $p^{k-m}=1$, or $k=m$, a contradiction.

So we now have $a!+p=p^{k}$. Let’s assume that $2p≤a$. Then $p,2p≤a$, therefore $p^{2}|a!$. But then $p^{2}|p^{k}-a!=p$, a contradiction. So $p≤a<2p.$ Now working with our equation, we get:

$a!+p=p^{k}$

$(p-1)!(p+1)⋯(a-1)a+1=p^{k-1}$

$(p-2)!(p+1)⋯(a-1)a=p^{k-2}+p^{k-3}+⋯+p+1.$

From here on, let’s assume that $(p-1)>4$. Then we know that $(p-2)! \equiv 0 \pmod{(p-1)}$. We also have $p^r \equiv 1 \pmod{(p-1)}$ for all nonnegative $r$. Applying these,

$0 \equiv 1+1+⋯+1+1=k-1 \pmod{(p-1)}.$

So $(k-1)=n(p-1)$, or $k=n(p-1)+1$ for some integer $n$. We have $a!+p=p^{n(p-1)+1}$. Since $a<2p$, we can write:

$a!≤(2p-1)!=[(p-(p-1))(p+(p-1))]⋯[(p-2)(p+2)][(p-1)(p+1)]p$

$a!≤[p^2-(p-1)^2][p^2-(p-2)^2]⋯[p^2-2^2 ][p^2-1^2]p$

$a!<(p^2)(p^2)⋯(p^2)(p^2-1)p$

$a!<(p^2)^{p-2}(p^2-1)p$

$a!<p^{2(p-2)+1} (p^2-1).$

And so:

$p^{n(p-1)+1}=a!+p<p^{2(p-2)+1} (p^2-1)+p=p^{2(p-1)+1}-p^{2p-3}+p<p^{2(p-1)+1}.$

Since $p>4$, we thus have $n(p-1)+1<2(p-1)+1$, or $n<2$. Clearly, $n$ must be nonnegative. If $n=0$, then $k=1$, which is a contradiction. So $n=1$ and $a!+p=p^p$.

Now let $p-1=2^{c}x$ and $p+1=2^{d}y$, with $x$ and $y$ odd. Note either that $c$ or $d$, but not both, must equal 1. Furthermore, let $P_2(e)$ denote the power of 2 in the prime factorization of some integer $e$. For example, by our definitions, $P_2(p-1)=c$ and $P_2(p+1)=d$. Note that if we can write $e$ out as a product $e_1\cdot e_2\cdot \cdot \cdot e_i$, then we can write $P_2(e)=P_2(e_1\cdot e_2⋯e_i)=P_2(e_1)+P_2(e_2)+⋯+P_2(e_i)$.

Case 1: $c=1$. We then have:

$a!=p^p-p=p(p^{p-1}-1)=p(p^{2x}-1)=p(p^{x}-1)(p^{x}+1)$

$a!=p(p-1)(p^{x-1}+⋯+p^2+p+1)(p+1)(p^{x-1}-⋯+p^2-p+1)$

$P_2(a!)=0+1 + 0 + d + 0$

$P_2(a!)=d+1$

Case 2: $d=1$. We then have:

$a!=p^p-p=p(p^{p-1}-1)=p(p^{2^cx}-1)$

$a!=p(p^x-1)(p^x+1)(p^{2x}+1)(p^{2^2x}+1)⋯(p^{2^{c-1}x}+1)$

$P_2(a!)=0+c + 1 + 1 + 1 + ⋯ + 1$

$P_2(a!)=2c$

To generalize, we can say that $P_2(a!)=c(d+1)$ to cover both of these cases. We now directly give a bound for $P_2(a!)$:

$P_2(a!)=⌊a/2⌋+⌊a/2^2⌋+⋯+⌊a/2^z⌋,$

where $z$ is such that $2^z≤a<2^{z+1}$. Since $a≥2^z$,

$P_2(a!)≥2^{z-1}+2^{z-2}+⋯+1=2^z-1.$

Moving on, we have:

$2^cx=p-1<a<2^{z+1}$

$2^{c-1}-1<2^{c-1}x-1<2^z-1≤P_2(a!)=c(d+1)$

$2^{c-1}-1<c(d+1)$.

Also,

$2^dy=p+1≤a+1<2^{z+1}+1<2^{z+1}+2$

$2^{d-1}-2<2^{d-1}y-2<2^z-1≤P_2(a!)=c(d+1)$

$2^{d-1}-2<c(d+1)$.

In the case where $c=1$, we use the second result here, and simplify: $2^{d-1}-3<d$. The only $d$ for which this equation holds are $d=1,2,3$. Once again, we can’t have $d=1$, so here, $d=2$ or $3$. Likewise, in the case where $d=1$, using the first result and simplifying, we get $2^{c-1}-1<2c$. In this case, omitting the solution of $c=1$, we must have $c=2, 3$ or $4$. Since $c≤4$ and $d≤3$, but one of $c,d$ is 1, we have $P_2(a!)=c(d+1)≤8$. But then note that $P_2(12!)=10$, so we must then have $a≤11$. But then $p≤a≤11$, and $p=7,11$ are the only possibilities. But $a!=7^7-7$ and $a!=11^{11}-11$ have no solutions for $a$.

Our original assumption that $(p-1)>4$ must then be false, so we consider $a!+p=p^k$ for $p=2,3,5$ and $p≤a<2p.$ If $p=2$, then $a=2$ or $3$. We see that $2!+2=2^2$ and $3!+2=2^3$. Moving on, if $p=3$, then $a=3,4,$ or $5$. From here, we get $3!+3=3^2$ and $4!+3=3^3$, but $a=5$ has no solutions. If $p=5$, then $a=5,6,7,8,$ or $9$. From these, we have $5!+5=5^3$ and no other solutions. Thus, the only solutions $(a,p,k)$ are $(2,2,2), (3,2,3), (3,3,2), (4,3,3),$ and $(5,5,3).$