Math 126, Number Theory
Second Test Answers
April 2006
Scale. 85–100 A, 70–84 B, 50–69 C. Median 70.
Problem 1. On Pythagorean triples. [18] Recall that a
Pythagorean triple (x, y, z) consists of three positive integers
such that x2+y2=z2. Show that for any Pythagorean triple
at least one of xand yis divisible by 3. [Hint: (mod 3).]
The proof revolves around the fact that the only squares
modulo 3 are 0 and 1. Here’s one version of it.
Suppose that (x, y, z ) is a Pythagorean triple such that
neither xnor yis divisible by 3. Since x2+y2=z2, therefore
x2+y2≡z2(mod 3)
but neither xnor yis congruent to 0 (mod 3). Then both
xand yare congruent to ±1 (mod 3), hence their squares
x2and y2are congruent to 1 (mod 3), and so their sum
x2+y2≡2 (mod 3). But 2 is not a square modulo 3, so
26≡ z2(mod 3), a contradiction. Thus, no Pythagorean
triple (x, y, z) has neither xnor ydivisible by 3. q.e.d.
Problem 2. Yes/no. [16; 4 points each part] For each of
the following just write “yes” or “no”. No explanation is
needed unless it’s not clear which is correct.
a. Fermat’s theorem implies that for prime p, 2p≡
2 (mod p). Does the converse hold, that is, if 2p≡
2 (mod p), then is pprime?
No, pseudoprimes also have this property.
b. If function fis multiplicative, then does that imply that
f(80) = f(8)f(10)?
No, 8 and 10 are not relatively prime. In fact, Euler’s phi
function is multiplicative, but φ(80) = φ(16)φ(5) = 8·4 = 32
while φ(8) = 4 and φ(10) = 4.
c. Is the number 4926834923 is the sum of two squares?
No. It’s congruent to 3 modulo 4, but sums of two squares
modulo 4 can only be congruent to 0, 1, or 2 modulo 4.
d. Does the Chinese remainder theorem imply that the pair
of linear congruences
x≡7 (mod 16)
x≡13 (mod 10)
has a unique solution modulo 160?
No. 16 and 10 are not relatively prime.
Problem 3. [18] Find at least two positive solutions of
quadratic Diophantine equation
2x2+xy −y2= 35.
[Hint: factor the left side of the equation.]
Factoring the left side, we get
(2x−y)(x+y) = 35.
We need to find factorings of 35 so that xand yturn out to
be positive integers.
2x−y x +y x y
1 35 12 23
5 7 4 3
7 5 4 1
35 1 12 −11
The three positive solutions for (x, y) are (4,3), (4,1), and
(12,23).
Problem 4. [20] On order and primitive roots.
a. Compute ord19(7), the order of 7 modulo 19. [It’s small.]
72= 49 ≡11 (mod 19)
73≡11 ·7 = 77 ≡1 (mod 19)
Therefore, ord19(7) = 3.
b. Note that 82≡7 (mod 19), and 23≡8 (mod 19). What
does that say about ord19(8) and ord19 (2)?
Since 82≡7 and 73≡1, therefore 86≡1. Hence ord19(8)
divides 6.
In fact, you can quickly rule out the possibilities of
ord19(8) being 1, 2, or 3, so ord19 (8) = 6.
Likewise 23≡8 and 86≡1 so 218 ≡1, so ord19(2) divides
18.
It’s a little more work to show that ord19(2) actually
equals 18.
c. How many primitive roots modulo 19 are there?
1
