Problem 5

Find all triples

(a,b,p)

of positive integers with

p

prime and

a^p=b!+p

Step 4 of 6: Rule out

a>p

to conclude

a=p

Previous step Next step

a=pk,\ k<p\ \Rightarrow\ k\mid b!\ \text{but}\ k\nmid a^p-p

Detailed analysis

Since $p\le a$ and $p\mid b!+p$ so $p\mid b!$ , in particular $p\le b$ . If $a>p$ , write $a=pk$ with integer $k\ge2$ ; since $k<p\le a$ , $k$ appears as a factor in $b!$ (because $k<p\le b$ ), so $k\mid b!$ , yet $k\mid a^p$ while $k\nmid p$ (as $k<p$ and $p$ is prime), so $k\nmid a^p-p=b!$ , a contradiction. Hence $a=p$ exactly.