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MAT 200, Logic, Language and Proof, Fall 2015
Solution to
Practice Questions
Problem 1. Find all solutions to the following linear diophantine equations
(i) 165m + 250n = 15;
(ii) 26m + 39n = 52;
(iii) 33m + 6n = 14.
Sol.
(i) : We have (165, 250) = 5 which divides 15, thus the equation has solutions. Using the Euclidean algorithm, we obtain the solution m0 = −9, n0 =
6. It follows that all solutions are given by
250
m = −9 +
q = −9 + 50q,
5
165
n=6−
q = 6 − 33q,
5
for q ∈ Z.
(ii) : We have (26, 39) = 13 which divides 52, thus the equation has solutions.
Using the Euclidean algorithm, we obtain the solution m0 = −4, n0 = 4. It
follows that all solutions are given by
39
m = −4 + q = −4 + 3q,
13
26
n = 4 − q = 4 − 2q,
13
for q ∈ Z.
(iii) : We have (33, 6) = 3 which does not divide 14, thus the equation has
no solution.
Problem 2. Let a, b be positive integers. Prove that a/(a, b) and b/(a, b)
are coprime.
Sol. Let d = (a, b) and let c be a common divisor of a/d and b/d. Then
a/d = cq for some q ∈ Z, i.e. a = cdq so that cd divides a. On the other
hand, b/d = cp for some p ∈ Z, i.e. b = cdp so that cd divides b. We obtain
that cd is a common divisor of a and b, so it has to divide their greatest
common divisor, i.e. cd divides d. It easily follows that c = 1 or c = −1, as
required.
Problem 3. Let n ∈ N. Prove that if there are no non-zero integer solutions
to the equation
xn + y n = z n ,
1
2
then there are no non-zero rational solutions.
Sol. We prove the contrapositive.
Suppose that there is a non-zero rational solution to this equation, say x =
a1 /b1 , y = a2 /b2 , z = a3 /b3 . Then
an1
an2
an3
+
=
.
bn1
bn2
bn3
Multiplying on both sides by bn1 bn2 bn3 , we get
(a1 b2 b3 )n + (a2 b1 b3 )n = (a3 b1 b2 )n
and x = a1 b2 b3 , y = a2 b1 b3 , z = a3 b1 b2 is a non-zero integer solution.
Problem 4. Find all integers x such that
3x ≡ 15
mod 18.
Hint : This is equivalent to solving the linear diophantine equation 3x +
18n = 15.
Sol. The solution to the linear diophantine equation 3x + 18n = 15 is
x = −1 + 6q, n = 1 − q
(q ∈ Z)
and thus the solution to the congruence
3x ≡ 15
mod 18
is x = −1 + 6q for q ∈ Z.
Problem 5. Prove, for positive integers n, that 7 divides 6n + 1 if and only
if n is odd.
Since 6 ≡ −1 mod 7, we have 6n + 1 ≡ (−1)n + 1 mod 7, which is congruent to 0 modulo 7 if and only if n is odd. This proves the result.
Problem 6. What is the last digit of 32014 ? What about the last digit of
732014 ?
Sol. First note that finding the last digit of a number is the same as finding
its residue modulo 10. Now, since 34 ≡ 1 mod 10, we have
32014 = 34·503+2 = (34 )503 · 32 ≡ 1503 · 32
mod 10
which implies that
32014 ≡ 9
mod 10
and thus the last digit of 32014 is 9. Since 73 ≡ 3 mod 10, the last digit of
732014 is also 9.
3
Problem 7. Prove that for each positive integer n, there is a sequence of
n consecutive integers all of which are composite.
Hint : Consider (n + 1)! + 2, (n + 1)! + 3, . . . , (n + 1)! + n + 1.
Sol. Let j ∈ {2, 3, . . . , n+1}. Then clearly j divides (n+1)!, so that j divides
(n + 1)! + j. Since 2 ≤ j < (n + 1)! + j, this shows that (n + 1)! + j is composite. It follows that the numbers (n + 1)! + 2, (n + 1)! + 3, . . . , (n + 1)! + n + 1
form a sequence of n composite consecutive integers.
Problem 8. Prove that there are infinitely many prime numbers which are
congruent to 3 modulo 4.
Hint : Proceed as in the proof of Theorem 23.5.1, but consider m =
4p1 p2 · · · pn − 1.
Sol. Suppose for a contradiction that there are only finitely many prime
numbers which are congruent to 3 modulo 4, say p1 , p2 , . . . , pn . Consider
the number
m = 4p1 p2 · · · pn − 1.
Then m is congruent to 3 modulo 4. Let p be a prime number dividing
m. Then p is odd, and p cannot be one of the pj ’s. Indeed, if p = pj for
some j, then p would divide 4p1 p2 · · · pn . Since it also divides m, it would
have to divide 4p1 p2 · · · pn − m = 1, a contradiction. It follows that p is
not congruent to 3 modulo 4, so the only possibility is that it is congruent
to 1 modulo 4. In other words, every prime divisor of m is congruent to 1
modulo 4. Since m is equal to the product of all its prime divisors, we get
that m has to be congruent to 1 modulo 4 as well (note that the product of
numbers which are congruent to 1 modulo 4 is congruent to 1 modulo 4).
This contradicts the fact that m is congruent to 3 modulo 4.