Math 115 Professor K. A. Ribet
First Midterm Exam February 25, 1998
☞Answer question #2 and three other questions.
1(6 points).Find all solutions to the congruence x2≡pmod p2when pis a prime number.
There are no solutions. Indeed, if xsatisfies the congruence, then x2≡0 mod p. Thus pdivides x2, so
pdivides xand thus p2divides x2. Since we then have x2≡0 mod p2, we do not have x2≡pmod p2.
(Hensel’s lemma has nothing to do with this problem: it doesn’t apply, so it gives no information.)
2(9 points).Using the equation 7·529 −3·1234 = 1, find an integer xwhich satisfies the two congruences
x≡n123 mod 529
321 mod 1234 and an integer ysuch that 7y≡1mod 1234. (No need to simplify.)
This is the question that you were more or less promised. I would take xto b e 123 ·(−3) ·1234 + 321 ·7·529,
or 733317; you can reduce this mod 652786, getting 80531 instead. Take yto be 529.
3(7 points).Suppose that pis a prime number. Which of the p+ 2 numbers p+ 1
k(0≤k≤p+ 1) are
divisible by p?
This was a “frequently omitted” question, but it’s reasonably easy. First proof: The formula p+ 1
k=
(p+ 1)!
k!(p+ 1 −k)! displays p+ 1
kas a fraction whose numerator is divisible by pbut not by p2and whose
denominator is divisible by ponly when kor p+ 1 −kis one of the two numbers p,p+ 1. Hence the
coefficient is divisible by pfor all kexcept the values 0, 1, p,p+ 1. Among the p+ 2 numbers in question,
p−2 of them are divisible by p. Second proof: Think about Pascal’s triangle and the fact that all the
numbers in the pth row are divisible by p(except for the two 1’s at the ends).
4(7 points).Let pbe a prime and let nbe a non-negative integer. Suppose that ais an integer prime to p.
Show that b:= apnsatisfies b≡amod pand bp−1≡1mod pn+1.
If ais as in the problem, then ap−1≡1 mod pby Fermat’s little theorem. Thus ais a root mod pof
f(x) = xp−1−1. Since f0(a) is then prime to p,acan be refined in exactly one way to a root of f(x)
mod pn+1. In other words, there exists a unique bmod pn+1 which satisfies bp−1≡1 mod pn+1 and b≡a
mod p. The p oint of this problem is that you can actually exhibit a bthat works. Namely, using Fermat’s
congruence ap≡amod p, you easily prove apn
≡amod pby induction. Thus bis indeed the same as a
mod p. Also, ais prime to pn+1, so aφ(pn+1 )≡1 mod pn+1. Since φ(pn+1 ) = (p−1)pn, this yields bp−1≡1
mod pn+1.
5(6 points).Show that n4+n2+ 1 is composite for all n≥2.
I added this problem at the last minute, just taking it from the book (problem 33 on page 32). If you calculate
two or three of these numbers, you see that they are composite but are not systematically divisible by any
particular number. This suggests an algebraic factorization. The point turns out to be that n4+n2+ 1 =
(n4+ 2n2+ 1) −n2is a difference of two squares. It’s thus the product n2+1+nn2+ 1 −n. To prove
that this is a non-trivial factorization, you have to see that n2−n+1 is bigger than 1, but this is easy (graph
it, or something).