Download Solutions to Practice Problems

Solutions to Practice Problems
March 2015
1. Given n = pq and φ(n) = (p − 1)(q − 1), we find p and q as the roots of the quadratic
equation
(x − p)(x − q) = x2 − (n − φ(n) + 1)x + n = 0.
The roots are
p, q =
1
2
p
n−φ(n)+1 ± (n−φ(n)+1)2 − 4n .
2. There are φ(12) = 4 Dirichlet characters mod 12. A partial list of values is given by
n
χ0 (n)
χ1 (n)
χ2 (n)
χ3 (n)
1
1
1
1
1
5
1
−1
−1
1
7
1
1
−1
−1
11
1
−1
1
−1
13
1
1
1
1
17
1
−1
−1
1
19
1
1
−1
−1
and χi (n) = 1 for n = 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, . . . (where gcd(n, 12) > 1).
3. By the extended Euclidean Algorithm, find integers a, b such that am + k(p−1) = 1.
Let g ≡ ra mod p. By Fermat’s Little Theorem,
g m ≡ (ra )m ≡ ram 1k ≡ ram (rp−1 )k ≡ ram+k(p−1) = r1 = r mod p.
Here we are assuming that gcd(r, p) = 1; if, however, r is divisible by p, then one may
instead simply choose g = 0.
4. No; since x2 ≡ y 2 ≡ 0 or 1 mod 4 for all integers x, y, we have x2 − 3y 2 ≡ 0, 1 or 2
mod 4.
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5. The primitive Pythagorean triples containing 100 are (2499, 100, 2501) and (621, 100,
629). These can be found by any of our approaches to the classification of primitive
Pythagorean triples.
6. Euclid’s Lemma states that if a prime p divides a product ab of two integers a and
b, then p divides at least one of the factors a, b. To prove this, we may suppose that
p does not divide a, so that gcd(a, p) = 1. By Euclid’s Algorithm, it follows that
ra + sp = 1 for some integers r, s; but then p divides r(ab) + (sb)p = (ra + sp)b = b
since it divides both terms on the left side.
7. A prime p has 100 digits iff it satisfies 1099 < p < 10100 . Such primes lie in the interval
[1, 10100 ] but not in the interval [1, 1099 ]. The number of such primes is
π(10100 ) − π(1099 ) ≈
10100
ln(10100 )
−
1099
ln(1099 )
≈ 3.9 × 1097
by the Prime Number Theorem.
8. The nonzero elements of Fp , except for 1 and −1, occur in pairs {a, a−1 }. So in the
product
Y
a,
06=a∈Fp
all factors cancel (in pairs) except for the factors 1 and −1, leaving a value of −1 as
the final value of the product. (Of course when p = 2 there is really only one such
factor since −1 = 1 in this case; but our answer holds since the final product is once
again −1 = 1.)
9. The Dirichlet character mod 2 defined by
0, if n is even;
χ(n) =
1, if n is odd
has Dirichlet series with Euler factorization given by
1
1
1
1
1
+
+
+
+
+ ···
32s
52s ! 72s
92s
112s
Y
1
L(χ, s) = 1 +
=
1 − χ(p)
ps
1
1
1
1
=
× ···.
1 − 31s
1 − 51s
1 − 71s
1 − 111s
p
2
This is almost the same as the Euler factorization for the Riemann zeta-function; only
the Euler factor for the prime p = 2 is missing, so
1
1 − 21s
L(χ, s) = ζ(s).
Evaluating at s = 2 gives
4
3 L(χ, 2)
= ζ(2) =
π2
6
π2
8
Solving for the given series, we obtain L(χ, 2) =
.
.
10. Recall that n is perfect iff σ(n) = 2n. Here σ is a multiplicative function satisfying
σ(pr ) = 1 + p + p2 + · · · + pr−1 + pr =
1−p−r−1 r
1−p−1 p
<
pr
1
1− p
.
If n = pr for some odd prime p and r ≥ 1, then we require 2n = σ(n) <
2<
1
1−
1
p
≤
1
1−
1
3
=
n
1
1− p
so
3
,
2
a contradiction. If n = pr q s for distinct odd primes p, q with exponents r, s ≥ 1, then
we require
2n = 2pr q s = σ(pr )σ(q s ) <
so
2<
1−
1
p
1
1−
1
q
≤
1−
1
3
1
pr
qs
1
1− p 1− q1
1−
1
5
=
15
;
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again, this is impossible.
11. No; the only Mersenne prime with last decimal digit equal to 3 is M2 = 22 − 1 = 3.
Let Mp = 2p − 1 where p is prime. If p = 4k + 1 for some integer k ≥ 1, then
Mp = 24k+1 − 1 = 2(16)k − 1 ≡ 2·6 − 1 ≡ 1 mod 10;
whereas if p = 4k + 3 then
Mp = 24k+3 − 1 = 8(16)k − 1 ≡ 8·6 − 1 ≡ 7 mod 10.
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12. If a is relatively prime with 1105 = 5·13·17, then by Fermat’s Little Theorem
a1104 = (a4 )276 ≡ 1276 ≡ 1 mod 5;
a1104 = (a12 )92 ≡ 192 ≡ 1 mod 13; and
a1104 = (a12 )69 ≡ 169 ≡ 1 mod 17
so a ≡ 1 mod 1105. (This last conclusion follows from the Chinese Remainder Theorem; but it is a more direct consequence of the fact that a − 1 is divisible by three
primes 5, 13 and 17, so it is divisible by their product.)
13. By Fermat’s Little Theorem, 11(M −1)/2 ≡ (a2 )(M −1)/2 ≡ aM −1 ≡ 1 mod M , so
11(M +1)/2 ≡ 11(M −1)/2 ·11 ≡ 11 . One solution is given by a ≡ 11(M +1)/4 mod M .
(Since M + 1 is divisible by 4, the exponent (M + 1)/4 is an integer.) The complete
list of solutions is given by a ≡ ±11(M +1)/4 mod M .
14. Let p1 , p2 , . . . , pr be the distinct primes dividing n; then we must solve
n = 2φ(n) = 2n 1 −
1
p1
1−
1
p2
··· 1 −
1
pr
,
i.e.
1=2 1−
1
p1
1−
1
p2
··· 1 −
1
pr
.
If n has any odd prime divisors pi , then the right hand side is never an integer (its
denominator will be divisible by pi ). So we must have n = 2k where k ≥ 1.
15. (a) The primes having at most 50 decimal digits lie in the interval [1, 1050 ]. Their
number is
1050
47
π(1050 ) ≈ ln(10
50 ) ≈ 8.7 × 10
by the Prime Number Theorem.
(b) The primes having exactly 50 decimal digits lie in the interval [1049 , 1050 ]. Their
number is
1050
1050
47
π(1050 ) − π(1049 ) ≈ ln(10
.
50 ) − ln(1050 ) ≈ 7.8 × 10
(c) Large primes end in either 1, 3, 7 or 9, in roughly equal numbers; so the number
of 50-digit primes ending in ‘3’ is roughly
1
4
π(1050 ) − π(1049 ) ≈ 1.9 × 1047 .
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(d) Large primes end in either X1, X3, X7 or X9, in roughly equal numbers, where
X represents any decimal digit; so the number of 50-digit primes ending in ‘33’ is
roughly
1
40
π(1050 ) − π(1049 ) ≈ 1.9 × 1046 .
16. By the extended Euclidean algorithm (or, in this simple case, by quick inspection) we
find 1·11 − 2·5 = 1, so the inverse of 5 mod 11 is −2 ≡ 9 and
5x ≡ 9 mod 11 ⇐⇒ x ≡ 9·9 ≡ 4 mod 11.
Write x = 11k + 4. Now we must solve
6x ≡ 5k + 24 ≡ 5 mod 61,
i.e. 5k ≡ 42 mod 61.
Once again, 61 − 12·5 = 1 so the inverse of 5 mod 61 is −12 ≡ 49 and
k ≡ 49·42 ≡ 45 mod 61.
Writing k = 61t + 45, this gives x = 11k + 4 = 671t + 499. In other words, the
simultaneous solutions of the original pair of linear congruences are given by x ≡ 499
mod 671.
17. No, it is not possible for f (n) to be prime for every positive integer n. Since f (x)
is nonconstant, it is either increasing or decreasing for all sufficiently large x. In the
decreasing case, f (x) is eventually negative for large x, and therefore not prime; so
we may assume there is a positive integer n such that f (x) is increasing for x ≥ n. By
assumption, p = f (n) is prime. Also f (n+p) ≡ f (n) ≡ p ≡ 0 mod p, i.e. f (n+p) is
divisible by p. Since f (n+p) is also prime, we must have f (n+p) = p. This contradicts
f (n+p) > f (n).
18. Yes. If p is composite, then there exists a ∈ {2, 3, . . . , p − 1} dividing p; but then
gcd(ap−1 , p) ≥ a > 1. (Note: Carmichael numbers are no exception to this fact. In
order for n to be a Carmichael number, one only requires an−1 ≡ 1 mod n for every
a relatively prime to n.)
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Solutions to Practice True/False Questions
1
F
2
F
3
F
4
F
5
T
6
F
7
T
8
T
9
F
10
F
Although you are not required to explain your answers to True/False questions, the following remarks may help to understand the solution key.
1. This special case of Fermat’s Last Theorem was proved many decades before the full
theorem was finally proved.
2. Fermat’s Little Theorem is used to verify that certain numbers are composite; but it
is of no direct value in finding prime factors of integers.
3. N 2 − 1 = (N + 1)(N − 1) is prime only for N ∈ {−2, 2}.
4. A counterexample is given by 4·2·5 − 1 = 3·13.
5. By Dirichlet’s Theorem, there are infinitely many primes congruent to 7 mod 8. None
of these are expressible as a sum of three squares.
6. Consider that 6 divides 4·9 = 36, but 6 divides neither 4 nor 9. (This shows that in
the statement of Euclid’s Lemma, the hypothesis that p is prime is essential.)
7. This fact, which is easily proved using the Fundamental Theorem of Arithmetic, was
used in our classification of primitive Pythagorean triples.
8. Recall that σ(p) is the sum of all positive divisors of p, including 1 and p itself; thus
σ(p) = 1 + p + sum of divisors of p between 1 and p . The latter sum is zero iff p is
prime.
9. The only prime p for which p2 + 1 is prime, is p = 2. All larger values of the prime p
yield an even value of p2 + 1 ≥ 10.
10. A counterexample is given by r = s = a = b = 1. From the relation ra + sb = d, one
can conlcude is that gcd(a, b) divides d, not that gcd(a, b) = d.
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