700 Number Theory
760.Note that
n= 1 + 10 +···+ 10 p−^2 =10 p−^1 − 1
10 − 1.
By Fermat’s little theorem the numerator is divisible byp, while the denominator is not.
Hence the conclusion.
761.We have the factorization
16320 = 26 · 3 · 5 · 17.First, note thatpab− 1 =(pa)b−1 is divisible bypa−1. Hencep^32 −1 is divisible by
p^2 −1,p^4 −1, andp^16 −1. By Fermat’s little theorem,p^2 − 1 =p^3 −^1 −1 is divisible
by 3,p^4 − 1 =p^5 −^1 −1 is divisible by 5, andp^16 − 1 =p^17 −^1 −1 is divisible by 17.
Here we used the fact thatp, being prime and greater than 17, is coprime to 3, 5 ,and 17.
We are left to show thatp^32 −1 is divisible by 2^6. Of course,pis odd, sayp= 2 m+1,
man integer. Thenp^32 − 1 =( 2 m+ 1 )^32 −1. Expanding with Newton’s binomial formula,
we get
( 2 m)^32 +(
32
1
)
( 2 m)^31 +···+(
32
2
)
( 2 m)^2 +(
32
1
)
( 2 m).In this sum all but the last five terms contain a power of two greater than or equal to 6.
On the other hand, it is easy to check that in
(
32
5
)
( 2 m)^5 +(
132
4
)
( 2 m)^4 +(
32
3
)
( 2 m)^3 +(
32
2
)
( 2 m)^2 +(
32
1
)
( 2 m)the first binomial coefficient is divisible by 2, the second by 2^2 , the third by 2^3 , the fourth
by 2^4 , and the fifth by 2^5. So this sum is divisible by 2^6 , and hence( 2 m+ 1 )^32 − 1 =p^32 − 1
is itself divisible by 2^6. This completes the solution.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by I. Tomescu)
762.Ifxis a solution to the equation from the statement, then using Fermat’s little
theorem, we obtain
1 ≡xp−^1 ≡ap− 1(^2) (modp).
Ifmis an integer, then every odd prime factorpofm^2 +1 must be of the form 4m+1,
withman integer. Indeed, in this case becausem^2 ≡− 1 (modp), and by what we just
proved,
(− 1 )
p− 21
= 1 ,
which means thatp−1 is divisible by 4.