Math 126, Number Theory
Second Test, alternate answers
11 Apr 2006
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 x,y, or zis divisible by 5. [Hint: what are
the squares mod 5?]
The squares modulo 5 are 0, 1, and −1 (which is the same
as 4). Thus, each of x2,y2and z2is one of those three. If
any one is congruent to 0 modulo 5, then it’s divisible by 5,
so one of x,y, or zis divisible by 5. That leaves the case
where each of x2,y2and z2is congruent to ±1. But the
sum of the first two is the third, and no combination of ±1
added to ±1 gives ±1 modulo 5. Thus, the remaining case
never occurs. Therefore, one of x,y, or zis divisible by 5.
q.e.d.
Problem 2. Yes/no. [16; 4 points each part]
a. Note that if (a, 15) = 1, then a4≡1 (mod 15). Also note
that φ(15) = 8. Does 15 have any primitive roots?
No. Since φ(15) = 8 a primitive root has order 8, but
since a4≡1 (mod 15), the highest order any totative can
be 4.
b. Fermat’s last theorem says that the Diophantine equa-
tions xn+yn=znhave no positive solutions for n > 2. Did
Fermat prove this theorem for any value of n > 2 at all?
Yes, and we studied his proof for n= 4.
c. If xy =z2and xand yare relatively prime, then does it
follow that each of xand yare perfect squares?
Yes, and we repeatedly used this principle to solve higher
order Diophantine equations.
d. If a4≡1 (mod n), then is the order of amodulo nequal
to 4?
No, it could be 1 or 2. For instance (−1)4≡1 (mod n),
but it’s order is not 4.
Problem 3. [18] Find at least one positive solution of
quadratic Diophantine equation
x2+xy −6y2= 21.
[Hint: factor the left side of the equation.]
The left side factors as (x+3y)(x−2y). We need to find a
factoring of 21 so that when we set the first factor to x+ 3y
and the second factor to x−2ywe get positive integers for
xand y. There are several factorings to consider. One that
works is x+ 3y= 21 and x−2y= 1. The solution to that
pair of equations is (x, y) = (9,4).
Problem 4. [15; 5 points each part] On order and primitive
roots.
a. What is the order of 2 modulo 17?
We need to raise 2 to higher and higher powers modulo
17 until we reach 1.
n1 2 3 4 5 6 7 8
2n24816151391
Thus, ord17 2 = 8.
b. Is 2 a primitive root modulo 17?
No, to be a primitive root, it would have to have an order
equal to φ(17) = 16.
c. How many primitive roots modulo 17 are there?
There are φ(16) = 8 of them.
Problem 5. [15] On Euler’s φfunction.
a. [5] How many positive integers less than 56 are rela-
tively prime to 56?
φ(56) = φ(8)φ(7) = 4 ·6 = 24
b. [10] Show that if n > 2 then 2|φ(n).
Here’s one proof. Let the prime decomposition of nbe
n=pe1
1pe2
2. . . pek
k.
Then
φ(n) = φ(pe1
1)φ(pe2
2). . . φ(pek
k).
If any one of the primes piis odd, then since
φ(pi) = (pi−1)pei−
i,
φ(pi) is even, and so φ(n) is even. Otherwise, there’s only
one prime p1= 2, so n= 2eis a power of 2. Now, since
n > 2, therefore e > 1, and φ(n) = φ(2e) = 2e−1is therefore
even. q.e.d.
Problem 6. [18] Solve the pair of linear congruences
4x+ 2y≡3 (mod 11)
2x−3y≡8 (mod 11)
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