Dr. Marques Sophie
Office 519
Number theory
Fall Semester 2013
marques@cims.nyu.edu
MIDTERM
Maximal Score: 200 points
Wednesday october 16th
Problem 1: (?) 50 points
1. Give all the congruence classes mod 99 solutions of the following system of congruences:
2x ≡ 1 mod 9
5x ≡ 3 mod11
2. Decide whenever the following linear diophantine equations have any solutions.
If so give the general solutions, if not says why there is no solutions
(a) 6x + 51y = 22;
(b) 162x + 64y = 4.
Solution:
1. 2 and 9 are coprime and 5 and 11 also and 9 and 11 are distinct prime number,
then we know by the CRT, that there is a unique congruence class solution for the
system mod 99. Let find a particular solution of the form x = 9z + 11y then we
get the
22y ≡ 1 mod 9
45z ≡ 3 mod11
This is equivalent to:
4y ≡ 1 mod 9
z ≡ 3 mod11
This is equivalent to:
−5y ≡ 1 mod 9
z
≡ 3 mod11
Then, obviously y = −2 and z = 3 are particular solutions. This gives us x =
9 × 3 + 11 × (−2) = 27 − 22 = 5 as particular solution. The unique congruence
class solution mod 99 is the one of 5.
2. (a) gcd(6, 51) = 3 - 22 so the equation has no solution.
1
(b) Applying the Euclidean algorithm to 81 and 32,
81
= 32 × 2 + 17
32
= 17 + 15
17
= 15 + 2
15 = 2 × 7 + 1
2=2×1+0
we get that gcd(162, 64) = 2|4. Thus the diophantine equation has infinitely
many solutions.
2 = 17 − 15
= 17 − (32 + 17)
= 17 × 2 − 32
= 2 × (81 − 32 × 2) − 32
= 2 × 81 − 5 × 32
Then 2 = 81 × 2 + (−5) × 32. Thus, x0 = 4 and y0 = −10 is a particular
solution.
Let (x, y) be another solution, then 81x + 32y = 2 = 81x0 + 32y0 , and
81(x − x0 ) = 32(y0 − y).Finally since gcd(81, 32) = 1 then 81|y0 − y. Then
there is k an integer such that 81k = y0 − y but then x − x0 = 32k. Finally,
the general solutions are of the form
x = 32k + 4
y = −81k − 10
Problem 2: (??) 45 points (The following questions are independents.)
1. Prove that the fraction
2. Prove that
n3 +2n
3
21n+4
14n+3
is irreducible, for any n ∈ Z.
is an integer, for any n ∈ Z.
3. Prove that if 3|a2 + b2 then 3|a and 3|b.
Solution:
1. We notice that −2(21n+4)+3(14n+3) = −8+9 = 1. Then, by Bezout’s theorem,
21n + 4 and 14n + 3 are coprime, and the fraction is irreducible.
2. The problem is equivalent to show that 3|n3 +2n or in congruence terms, n3 +2n ≡
0 mod 3. There are three cases to observe:
n ≡ 0 mod 3 then n3 + 2n ≡ 0 mod 3
n ≡ 1 mod 3 then n3 + 2n ≡ 3 ≡ 0 mod 3
n ≡ 2 mod 3 then n3 + 2n ≡ 8 + 4 ≡ 0 mod 3
Then in all possible cases,
n3 +2n
3
is an integer.
2
3. Suppose by the contrapositive that one of a or b is not divisible by 3. Let say that
a is not divisible by 3, the other case will be completely similar. Then, a ≡ 1 or
2 mod 3 in both case a2 ≡ 1 mod 3, then there are 3 possibles cases
b ≡ 0 mod 3 then a2 + b2 ≡ 1 6≡ 0 mod 3
b ≡ 1 mod 3 then a2 + b2 ≡ 2 6≡ 0 mod 3
b ≡ 2 mod 3 then a2 + b2 ≡ 2 6≡ 0 mod 3
Then in the three cases, 3 - a2 + b2 .
Problem 3: (?) 10 points
What is the remainder of 52012 when divided by 11?
Solution: Since 11 is prime, we can apply the Fermat’s theorem, observing that 2013 =
201 × 10 + 3, then
52013 ≡ 5201×10+3 ≡ 52 ≡ 3 mod 11
Problem 4: (?) 10 points
Suppose that a, b and c are positive integers with b|c and m = [a, c]. Prove that
[a, b] ≤ m.
Solution: Since b|c and c|[a, c] then b|[a, c] and a|[a, c] by definition of the L.C.M.
Then again by definition of L.C.M, [b, a] ≤ [a, c]
Problem 5: (?) 50 points
Use induction to show that
n
X
i(i + 1) =
i=1
n(n + 1)(n + 2)
3
Solution: We start the induction for n = 1, then
1×2=
1×2×3
=2
3
Suppose that the property is true for some n and we want to prove it for n + 1.
We have
Pn+1
Pn
i=1 i(i + 1) =
i=1 i(i + 1) + (n + 1)(n + 2)
n(n+1)(n+2)
=
+ (n + 1)(n + 2)
3
n(n+1)(n+2)
=
+ 3(n+1)(n+2)
3
3
= (n+1)(n+2)(n+3)
3
Then, we have proven by induction that the property is true.
Exercise 6: (??) 35 points
1. Prove that if a ≡ b mod p, for every prime p, then a = b.
3
2. Let a, m, n, b, c ∈ N such that (m, n) = 1. Suppose that
b
a ≡ 1 mod m
ac ≡ 1 mod n
Prove that
abc ≡ 1 mod mn
Solution:
1. Without loss of generality, we can suppose that a − b is positive, take p larger than
a − b then a − b ≡ 0 mod p by assumption. But p|(a − b) and 0 ≤ a − b < p, so
the only possibility is that a = b.
2. Since ab ≡ 1 mod m then abc = (ab )c ≡ 1mod m and ac ≡ 1 mod n then
abc = (ac )b ≡ 1mod n. Then since gcd(n, m) = 1. Then,
abc ≡ 1 mod mn
1
1
(?) = easy , (??)= medium, (???)= challenge
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