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```Nipissing University
Department of Computer Science & Mathematics
MATH 3126 - Number Theory
Instructor: Dr. Tzvetalin S. Vassilev
Time: 90 minutes
Midterm Examination (Solution)
October 25, 2011
CLOSED BOOK. NO CALCULATORS. SHOW YOUR WORK
1. [5 pts.] Let tn denote the n-th triangular number. Show that t608 is a product of
three consecutive integers.
Solution: Since tn = n(n+1)
, t608 = 608·609
= 304 · 609 = 2 · 152 · 3 · 203 = 2 · 2 · 76 · 3 ·
2
2
7·29 = 2·2·2·38·3·7·29 = 2·2·2·2·19·3·7·29 = (2·2·2·7)·(3·19)·(2·29) = 56·57·58
Please note: no calculator or multiplication of large numbers is required to solve
this question. As a matter of fact, any of these is detrimental to solving the
question.
2. [5 pts.] Find a formula for an , n ≥ 0, the n-th term of the sequence
3, 10, 21, 36, 55, 78, . . .
Solution: Using the finite difference method, we have:
3 10 21 36 55 78
7 11 15 19 23
4 4 4 4
n
n
n
Thus, an = 3 0 + 7 1 + 4 2 = 3 + 7n + 2n(n − 1) = 2n2 + 5n + 3.
3. [5 pts.] Let un denote the n-th Fibonacci number. Prove that
u2n+1 − u2n = un−1 un+2
Solution: Direct proof is available here. Those interested in pursuing an inductive
approach may do so.
u2n+1 − u2n = (un+1 − un )(un+1 + un ) = un−1 un+2
4. [10 pts.] Let tn denote the n-th triangular number. Find the sum of the reciprocals
of the triangular numbers, i.e.
∞
X
1
1 1
1
=1+ + +
+ ...
t
3
6
10
n
n=1
Solution: Having in mind that
k
k
k
X
X
X
1
2
1
k
1
=
=2
=2
=2 1−
t
n(n + 1)
n(n + 1)
k+1
k+1
n=1 n
n=1
n=1
∞
k
X
X
1
1
1
= lim
= lim 2 1 −
=2
k→∞
k→∞
t
t
k+1
n=1 n
n=1 n
5. [15 pts.] Let tn denote the n-th triangular number. Prove that tm+n = tm +tn +mn.
Solution: Recall that tn = n(n+1)
. Thus,
2
n(n + 1) n(n + 1)
m2 + m n2 + n 2mn
+
+ mn =
+
+
=
2
2
2
2
2
m2 + n2 + 2mn + m + n
(m + n)2 + (m + n)
(m + n)(m + n + 1)
=
=
=
= tm+n
2
2
2
tm + tn + mn =
6. [5 pts.] Prove that 7 divides 32n+1 + 2n+2 , for any positive integer n.
Solution: We will give an inductive proof here.
Base case: n = 1. 32·1+1 + 21+2 = 33 + 23 = 27 + 8 = 35. 7 divides 35, which
establishes the truth of the claim for n = 1.
Inductive step: Assume that for some integer k ≥ 1, 7|32k+1 + 2k+2 .
We have to prove that 7|32(k+1)+1 + 2(k+1)+2 = 32k+3 + 2k+3 .
Further, 32k+3 + 2k+3 = 32 · 32k+1 + 2 · 2k+2 = 9 · 32k+1 + 2 · 2k+2 = 7 · 32k+1 +
2 32k+1 + 2k+2 .
But, 7|32k+1 + 2k+2 according to the inductive hypothesis.
Also, 7|7 · 32k+1 by definition.
Thus, 7|32(k+1)+1 + 2(k+1)+2 = 32k+3 + 2k+3 , which concludes the inductive step.
Conclusion: By mathematical induction, ∀n ∈ Z+ , 7|32n+1 + 2n+2 .
7. [5 pts.] Prove that there cannot exist integers a and b such that gcd(a, b) = 3 and
a + b = 65.
Solution: We know that gcd(a, b) divides any linear combination of a and b. Thus,
here 3 has to divide 65, which is absurd. Therefore, no such numbers exist.
8. [5 pts.] If a divides b, determine gcd(a, b) and lcm(a, b).
Solution: Obviously, gcd(a, b) = a and lcm(a, b) = b.
9. [5 pts.] Give an example to show that it is not necessarily the case that
gcd(a, b, c) · lcm(a, b, c) = abc
Solution: Take a = b = c = 2. Then gcd(a, b, c) = lcm(a, b, c) = 2, and therefore
gcd(a, b, c) · lcm(a, b, c) = 4, while abc = 8.
10. [5 pts.] Express gcd(481, 299) as a linear combination of 481 and 299.
Solution: Here we use the Saunderson’s Algorithm with 481 and 299:
1 · 481 − 0 · 299 = 481
0 · 481 − 1 · 299 = −299
1 · 481 − 1 · 299 = 182
1 · 481 − 2 · 299 = −117
2 · 481 − 3 · 299 = 65
3 · 481 − 5 · 299 = −52
5 · 481 − 8 · 299 = 13
Since gcd(481, 299) = 13 the task is accomplished, 13 = 5 · 481 − 8 · 299.
11. [10 pts.] Show that in any set of (n + 1) different integers selected from the set
{1, 2, . . . , 2n} there must exist a pair of coprime integers.
Solution: Here we use the Dirichlet Principle. Consider the n pairs of coprime
integers (1, 2), (3, 4), (5, 6), . . . , (2n − 1, 2n). Together these pairs comprise the set
S = {1, 2, . . . , 2n}. If we select (n + 1) different numbers from S, then by Dirichlet
Principle we are guaranteed that we have selected both elements of at least one of
the n pairs. Thus, we have selected a pair of coprime numbers.
12. [10 pts.] Prove, for all positive integers n, that
n5 n3 7n
+
+
5
3
15
is an integer.
Solution: Another inductive proof here.
5
3
7
3
5
7
Base case: n = 1. 15 + 13 + 7·1
= 51 + 13 + 15
= 15
+ 15
+ 15
= 1. Thus, the base
15
case is verified.
5
3
Inductive step: Assume that for some integer k ≥ 1, k5 + k3 + 7k
is an integer.
15
(k+1)5
(k+1)3
7(k+1)
We have to show that 5 + 3 + 15 is an integer.
k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 k 3 + 3k 2 + 3k + 1 7k + 7
(k + 1)5 (k + 1)3 7(k + 1)
+
+
=
+
+
=
5
3
15
5
3
15
5
k
k 3 7k
7
1 1
4
3
2
2
=
+
+
+ +
+ (k + 2k + 2k + k) + (k + k) +
3
5
15
3 5 15
The expression in the first set of brackets is an integer, according to the inductive
hypothesis. The expressions in the second and third set of brackets are integers
since k is an integer (i.e. products and sums of integers are integers), the expression
in the fourth set of brackets is an integer, too (it equals 1, as shown in the base
case). Thus, the entire expression is an integer, being represented as a sum of four
integers.
5
3
Conclusion: By mathematical induction, ∀n ∈ Z+ , n5 + n3 + 7n
is an integer.
15
13. [15 pts.] Show that if y 2 = x3 + 2, then both x and y are odd.
Solution: Assume x is even, that is x = 2k for some integer k. Then, y 2 = x3 +2 =
(2k)3 + 2 = 8k 3 + 2 = 4(2k 3 ) + 2. Thus, y 2 has a remainder of 2 when divided by
4. This is absurd. Therefore, x cannot be even, i.e. x is odd. Since x is odd, x3 is
odd, and subsequently x3 + 2 is odd. Then y 2 is odd, and y is odd.
```
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