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Math 126, Number Theory
Final Answers
May 2006
Scale. 85–100 A. 70-84 B. 55 69 C. Median 86.
Problem 1. [20] Carefully prove the following statement.
If x = r/s is a rational solution of the equation
anxn^ + an− 1 xn−^1 + · · · + a 1 x + a 0 = 0
where all the coefficients an, an− 1 ,... , a 1 , a 0 are in- tegers, and if r and s are relatively prime, then r divides the constant term a 0 while s divides the leading coefficient an.
Naturally, there are many proofs. Here’s a straightforward one.
Proof: Let x = r/x be a solution with r and s relatively prime. Then
an(r/s)n^ + an− 1 (r/s)n−^1 + · · · + a 1 (r/s) + a 0 = 0.
Multiplying each side of the equation by sn, we have
anrn^ + an− 1 rn−^1 s + · · · + a 1 rsn−^1 + a 0 sn^ = 0.
All but one of the terms contains a factor of r, so, moving the remaining term to the other side, we can rewrite this equation as
(anrn−^1 + an− 1 rn−^2 s + · · · + a 1 sn−^1 )r = −a 0 sn.
Since r divides the left hand side of the equation (by the definition of divisibility), therefore r divides the right hand side −a 0 sn. But r is relatively prime to s, therefore r is relatively prime to sn. But r divides −a 0 sn, hence r divides −a 0 , and thus r divides a 0. Likewise, all but one of the terms contains a factor of s, so we can rewrite the equation as
(an− 1 rn−^1 + · · · + a 1 rsn−^2 + a 0 sn−^1 )s = −anrn.
Since s divides the left hand side, it also divides −anrn. But s being relatively prime to r, it is also relatively prime to rn. Hence s divides −an, and so also divides an. q.e.d.
Problem 2. [24; 6 points each part] The Euclidean algo- rithm has been essential for us this semester.
a. Choose two of the following numbers, call them m and n, and show how the Euclidean algorithm is used to find their greatest common divisor d = (m, n).
2907 , 3128 , 4807 , 9775
Each choice of m and n leads to different greatest common divisors. Let’s take m = 4807 and n = 3128. Then
4807 = 3128 + 1679 3128 = 1679 + 1449 1679 = 1449 + 230 1449 = 6 · 230 + 69 230 = 3 · 69 + 23 69 = 3 · 23
Therefore, the greatest common divisor d is 23.
b. Use the work you did in part a to show how d is a linear combination of m and n, that is, solve the linear Diophantine equation mx + ny = d. One way to do that is to work the equations in part a backwards.
23 = 230 − 3 · 69 = 230 − 3 · (1449 − 6 · 230) = 19 · 230 − 3 · 1449 = 19 · (1679 − 1449) − 3 · 1449 = 19 · 1679 − 22 · 1449 = 19 · 1679 − 22 · (3128 − 1679) = 41 · 1679 − 22 · 3128 = 41 · (4807 − 3128) − 22 · 3128 = 41 · 4807 − 63 · 3128
c. Continuing with this example, show how you can use your results in part b to solve the linear congruence mx ≡ 6 d (mod n). That linear congruence is equivalent to the linear Dio- phantine equation
mx + ny = 6d.
Since d = 41m − 63 n, therefore 6d = 6 · 41 m − 6 · 63 n = 246 m − 378 n. Thus, x ≡ 246 (mod n) is a solution.
d. The Euclidean algorithm also is used to find continued fraction expansions. Find the continued fraction expansion of m/n, and show your work.
You can find this as a restatement of part a. 4807 3128
Alternatively, you can read the coefficients 1, 1, 6, 3, 3 di- rectly from the equations in part a.
Problem 3. [18] Consider the system of three congruences
x ≡ 4 (mod 11) x ≡ 3 (mod 9) x ≡ 3 (mod 10)
Since the moduli are pairwise relatively prime, the Chinese remainder theorem says that there is a unique solution to this system modulo 990, the product of the moduli. Find that solution and show your work.
There are a couple of ways to do this. One is to take the first two congruences and replace them by a single congru- ence modulo 99, then take that congruence along with the third to get a single congruence modulo 990. Here’s a different method that uses all three together. Step 1. For each modulus, find a reciprocal of the product of the remaining moduli modulo the given modulus. For the first modulus, 11, that means we need the reciprocal of 90 modulo 11, that is, we need to solve
90 y ≡ 1 (mod 11).
That’s the same as 2y ≡ 1 (mod 11), and that can be easily found by searching to be y ≡ 6 (mod 11). Thus, 6 is the reciprocal we’re looking for. For the second modulus, 9, we need the reciprocal of 110 modulo 9. That’s the same as the reciprocal of 2 modulo 9, which is 5.
For the third modulus, 10, we need the reciprocal of 99 modulo 10. That’s the same as the reciprocal of 9 modulo 10, which is 9. Step 2. To get x sum three products abc, one for each congruence, where a is the constant in the congruence, b is the product of the other moduli, and c is the reciprocal found in the previous step. That gives us
4 · 90 · 6 + 3 · 110 · 5 + 3 · 99 · 9 = 2160 + 1650 + 2673 = 6483
and then reduce this number modulo the product 990 of all three moduli. That gives a final answer of x ≡ 543 (mod 990).
Problem 4. [18] (page 107, exercise 3) Suppose that (a, n) = 1. Prove that
ab^ ≡ ac^ (mod n)
if and only if b ≡ c (mod ordn(a)). (You may use the theorems in the text, of course.) Of course, there are many proofs. The key theorem 3. is the one that says
ax^ ≡ a (mod n) iff x ≡ 1 (mod ordn(a)).
Here’s a proof that uses this theorem in both directions. Proof ⇒: Let ab^ ≡ ac^ (mod n). Suppose first that b ≥ c. Then the congruence ab^ − ac^ ≡ 0 (mod n) can be written as ac(ab−c^ − 1) ≡ 0 (mod n). Thus, n|ac(ab−c^ − 1). But (a, n) = 1, so (ac, n) = 1, and therefore n|(ab−c^ − 1). That says ab−c^ ≡ 1 (mod n). Therefore, by theorem 3.26, b ≡ c (mod ordn(a)). Thus, we’ve shown that if b ≥ c, then b ≡ c (mod ordn(a)) as required. But if c ≥ b, the same argument holds with b and c interchanged in the argument. Thus, in all cases b ≡ c (mod ordn(a)). Proof ⇐: Let b ≡ c (mod ordn(a)). Again, first take the case that b ≥ c. Then b = c+k ordn(a) for some nonnegative integer k. Therefore, modulo n,
ab^ ≡ ac+k^ ordn(a) ≡ ac(aordn(a))k ≡ ac 1 k ≡ ac
Likewise, if c ≥ b, we can also show ab^ ≡ ac^ (mod n). q.e.d.
Problem 5. [20; 10 points each part] Consider the Pell equation x^2 − 30 y^2 = 1. a. Find solution to the equation from the continued frac- tion expansion of
30, which is √ 30 = 5 +
