Solution 45 (almost complete)
pk+pl+pm=n2Clearly we cannot have
k=l=m because this would imply that
n had a factor of
9 and hence that
3∣p. Furthermore, ... Therefore, without loss of generality we can assume
k<m,l. By writing
n2=pkq2 and dividing, we get:
1+pl−k+pm−k=q2⇒pl−k+pm−k=(q+1)(q−1)Since
p is odd, the right hand side is even and is a multiple of
23. Therefore in the case where
l=m (with equality
4 does not divide the right hand side, so this is impossible) where we can assume without loss of generality that
m−l≥1,
pm−l=−1mod8.
It can be shown by exhaustion that
pm−l=−1mod8 for some
m−l≥1 if, and only if,
p=−1mod8, completing the proof.