Mathematics 2500
Elementary Number Theory
2011 Assignment 1 Solutions
Updated February 24, 2011
1:6 Find two different solutions of 299x + 247y = 13.
Solution: Using the matrix method, 1 0 299 0 1 247 ≡ 1 −1 52 0 1 247 ≡ 1 −1 52 −5 6 −13 ≡ −19 23 0 −5 6 −13 ≡ 5 −6 13 −19 23 0 .
So (x, y) = (5, −6) is one solution. For a second solution add the second row to the first in the final matrix: −14 17 13
−19 23 0
.. Since this is also in the “target” form, (x, y) = (−14, 17) is another solution.
Comments: There are infinitely many solutions, so our answers are not unique. There are also many different ways to navigate through the steps of the matrix reduction, but I generally prefer at each point to take the step producing a number whose absolute value is as small as possible in the right hand column. Note that the instruction I gave you calls for the matrix method. You will lose marks if you did not follow this instruction. You could also follow the Euclidean algorithm, which I believe is one step longer in this problem. Whether you use the steps of the Euclidean algorithm or an approach like mine. It is not necessary to explicitly write down the general solution; indeed, this was not covered in Section 1 and wouldn’t really be appropriate.
1:9 Prove that ((a, b), b) = (a, b). Solution: Let c = (a, b) and d = ((a, b), b).
Thus, d = (c, b). Now obviously c|c and, by definition, c|b, so c ≤ d.
Similarly, d|c because d = (c, b), so d ≤ c (greatest common divisors are positive numbers, so d|c implies the inequality).
Combining the two inequalities gives c = d, as required. 2
Comments: Most nice proofs will involve two inequalities like this.
2:2 Find the prime-power decompositions of 2345, 45670, and 999999999999. (Note that 101|1000001.) Solution: 2345 = 5 · 7 · 67 45670 = 2 · 5 · 4567 999999999999 = 999999 · 1000001 = 32· 111111 · 101 · 991 = 32· 111 · 1001 · 101 · 991 = 33 · 7 · 37 · 101 · 143 · 991 = 33· 7 · 11 · 13 · 37 · 101 · 991.
Comments: If you tried to do all of these on your calculator, you did too much work. Remember that Appendix C is a table of odd numbers up to 10,000 either indicating that they are prime or showing their first prime factor. When you get used to using this table you’ll find exercises like this fairly mechanical and easy.
2:4 (a) De Bouvelles (1509) claimed that one or both of 6n + 1 and 6n − 1 are primes for all n ≥ 1. Show that he was wrong.
(b) Show that there are infinitely many n such that both 6n + 1 and 6n − 1 are composite.
Solution: You are asked only to do (b) and only the case 6n − 1 —that is, show that there are infinitely many composition numbers of this form.
The standard factorization ae − 1 = (a − 1)(ae−1 + ae−2 + ae−3+ · · · + a2+ a + 1)
holds for all a and all e ∈ Z+. Thus it holds for a = 6. Letting e run over the positive
integers greater than 1 we obtain infinitely many numbers greater than 5 and divisible by a − 1 = 5, as required.
Comments: It is easy to solve both parts simultaneously by letting e run over only the odd positive integers greater than 1. Recall that another standard factorization has ae + 1 = (a + 1)(ae−1 − ae−2 + ae−3− · · · ± 1), when e is odd. Thus ae has the
form 6n, where 5|(6n − 1) and 7|(6n + 1).
2:5, 6 Prove that if n is a square, then each exponent in its prime-power decomposition is even.
and
Prove that if each exponent in the prime-power decomposition of n is even, then n is s square.
Solution: Let n = m2, where m has prime-power decomposition m = pe1
1 · · · p ek k . Then n = m2 = (pe1 1 · · · p ek k ) 2 = p2e1 1 · · · p 2ek
k . By the fundamental Theorem of Arithmetic, the
prime-power decomposition of n is unique, so this is it—and every exponent is even, as required. This proves #5.
Now suppose, conversely, that n = p2e1
1 · · · p 2ek k . Then n = p 2e1 1 · · · p ek k 2 —a square, as required. This proves #6.
Comments: You must invoke uniqueness in #5 else one might say, “yes, it’s all well and good that you have some prime power decomposition with all even powers, but the prime power decomposition of n might be a different decomposition, with odd powers.”
2:14 Prove that if n is composite, 2n− 1 is composite.
Solution: Suppose n = ab, where a, b > 1. Then n = (2a)b
= (2a− 1)(2a)b−1+ (2a)b−2+ · · · + 1. Since a > 1 and b > 1, the two factors are both > 1, so n is composite.
Comments: This is probably the only approach that works.
3:2 Find all the integer solutions of 2x + y = 2, 3x − 4y = 0, and 15x + 18y = 17.
Solution: In the first two case (a, b) = 1 and a particular solution (x0, y0) is immediately obvious
so Theorem 1 gives an immediate solution, (x, y) = (x0+ bt, y0− at).
For the first, (x, y) = (1, 0) is a particular solution, so the general solution is (x, y) = (1 + t, −2t), t ∈ Z.
For the second, (x, y) = (4, 3) is a particular solution, so the general solution is (x, y) = (4 − 4t, 3 − 3t), t ∈ Z.
For the third we observe (or derive using Euclidean algorithm or the matrix method) that (a, b) = 3 6 | 17, so there is no solution.
Comments: It doesn’t hurt to explicitly write down a formula before using it. In fact, I generally consider showing a correct and appropriate formula (on exams, etc) to be worth one (part) mark even when everything else is wrong. Normally you should never use names without defining them unless they are used in a problem. In explaining this I used names a, b, x0 and y0 —improperly by this rule. This is
marginally acceptable in this case because these names are understood in context. You should write t ∈ Z for a similar reason. It is not good practice to assume your reader automatically understands the possible values of a parameter.
I write my solutions here as number pairs, “(x, y) = · · · ”. This is not strictly necessary and it’s perfectly acceptable to use pairs of equations “x = · · · , y = · · · ” instead.
3:4 Find all the solutions in positive integers of x + y = 2, 3x − 4y = 0, and 7x + 15y = 51 Solution: First solve over the integers.
Once more, the first is immediate, (x, y) = (1 + t, 1 − t), t ∈ Z. The third is easily solved with the matrix method, 1 0 7
0 1 15 ≡ 1 0 7 −2 1 1 ≡ 15 −7 0 −2 1 1 ≡ −2 1 1 15 −7 0 . So 7(−2) + 15(1) = 1. Multiplying by 51 gives the particular solution 7(−102) + 15(51) = 51, so the general solution is (x, y) = (−102 + 15t, 51 − 7t), t ∈ Z.
Then place bounds on t. In the first, 1 + t > 0, 1 − t > 0, so −1 < t < 1, i.e., t = 0. In the second, we obtain t < 1, so t = 0, −1, −2, . . .. In the third, −102 + 15t > 0, 51 − 7t > 0, so 10215 < t < 517, so t = 7.
Substituting t into the parametric solutions we obtain the answers, namely (a) (x, y) = (1, 1); (b) (x, y) = (4 − 4t, 3 − 3t), t < 1; and (c) (x, y) = (3, 2).
Comments: Much of the second part is a repeat of part of problem 2 so it is not marked here. The form of the solution is not unique (It never is when a parameter is involved). In particular, in the second and third parts it makes sense to use −t as a parameter instead of t, so your answers would look like (b) (x, y) = (4 + 4t, 3 + 3t), t ≥ 0; and (c) (x, y) = (−102 − 15t, 5 + 7t), t ∈ {0, 1, 2, 3, 4, 5, 6}. While in (b) it is necessary to simply specify in terms of t because there are infinitely many solutions whereas in (a) and (c) there was only one so t would be an excrescence. If bounds like 3 < t < 8 were found, you could opt for a parameterless list of the three solutions or parametric form.
3:8 The enrollment in a number theory class consists of sophomores, juniors, and backward seniors. If each sophomore contributes $1.25, each junior $.90, and each senior $.50, the instructor will have a fund of $25. There are 26 students; how many of each? Solution: Let a, b and c represent the numbers of sophomores, juniors and seniors in the class,
respectively. From the story problem we have a + b + c = 26 125a + 90b + 50c = 2500
Subtracting 50× the first equation from the second gives an equation in a and b: 75a + 40b = 1200
We solve over the integers with the matrix method: 1 0 75 0 1 40 ≡ 1 −2 −5 0 1 40 ≡ 1 −2 −5 8 −15 0 ≡ −1 2 5 8 −15 0 . The first row gives 75(−1)+40(2) = 5. Multiplying by 240 gives 74(−240)+40(480) = 1200, a particular solution to the equation. Using the information in the second line of the matrix yields the full solution, (a, b) = (−240 + 8t, 480 − 15t).
Since a+b+c = 26 we have c = 26−a−b = 26−(−240+8t)−(480−15t) = −214+7t. The wording suggests each group is represented, so 0 < a, b, c ≤ 26. In particular, this holds for b, so 0 < 480 − 15t ≤ 26, so 454 ≤ 15t < 480, which implies t = 31. So (a, b, c) = (−240 + 8 · 31, 480 − 15 · 31, −214 + 7 · 31) = (8, 15, 3). There are 8 juniors, 15 sophomores and 3 seniors.
Comments: I introduce names a, b and c in my solution—mandatory in this situation. Why? The names are not standard and can’t be inferred from context. You must tell the reader everything they need to know that is not already inferable from the problem itself or from general context. You should understand why this is so and do it automatically. I will let you get away with this here, but count on a free pass in future assignments or tests! a, b or c = 0 is not explicitly excluded; in this case there is another possible solution, (a, b, c) = (16, 0, 10). You may notice that much work is saved by dividing the second equation by 5. I avoided doing so here because we don’t assume that you immediately notice the greatest common denominator of the coefficients; the method works whether you observe such shortcuts or not.
4:2 Find the least residue of 1492 (mod 4), (mod 10) and (mod 101).
Solution: 1492 = 373 · 4 ≡ 0 (mod 4); 1492 = 149 · 10 + 2 ≡ 2 (mod 10); and 1492 = 14 · 101 + 78 ≡ 78 (mod 101). So the required least residues are 0, 2 and 78 respectively. Comments: Not marked. “Work” consists of using the division algorithm to express 1492 as in Theorem 1. Note, it suffices to do this in stages, or by observation. For example, you might note that 1500, being a multiple of 100, is a multiple of 4 so 1492 ≡ −8 (mod 4), and draw the conclusion from there. When a random need for a least residue arises among small numbers (say less than 10 digits) this kind of ad hoc approach is my preferred method. However, it is good practice to show how you use results from the section a problem is supposed to be “practice” for.
4:3 Prove or disprove that if a ≡ b (mod m), then a2 ≡ b2 (mod m)
Solution: This is an immediate consequence of Lemma 1 (c). (Take c = a and d = b.) Here is an alternate approach using the definition of congruence:
Let a ≡ b (mod m). Then m|(b − a) by definition. Thus, m|(b − a)(b + a). That is, m|(b2− a2. So a2 ≡ b2 (mod m), by definition.
Comments: Either method has its merits. I expect you to understand both ap-proaches. Lemma 1 is a bag of tools whose use should become like breathing to you. On the other hand, you should be so accustomed to the definition, it should be al-most automatic to construct the alternate proof, which has the aesthetic merit that it depends on nothing beyond the notion of divisibility and the definition of congruence.
4:6 Find all m such that 1848 ≡ 1914 (mod m).
Solution: 1848 ≡ 1914 (mod m) if and only if m|(1914 − 1848). That is, m is any positive divisor of 1914 − 1848 = 66 = 2 · 3 · 11. These are easy to list: 1, 2, 3, 6, 11, 22, 33, 66.
Comments: Not everyone admits 1 as a valid modulus. However, as discussed in class I see no harm in doing so, indeed I believe there is merit in it. I also, you will recall, admit 0 as a modulus, though that is not relevant to this question.
23:S1:2 (a) Prove that (a, b) = (a, c) = 1 implies (a, bc) = 1. (b) Prove that (a, b) = 1 and (c|a) imply (c, b) = 1.
Solution: (a) By Theorem XXX, there exist integers x, y, u, v such that ax + by = au + cv = 1. Multiplying, we obtain
1 = (ax + by)(au + cv) = a2xu + acxv + abyu + bcyv = a(axu + cxv + byu) + bc(yv). Now (a, bc)|a and (a, bc)|bc, so (a, bc)|[a(axu + cxv + byu) + bc(yv)].
That is, (a, bc)|1 — so (a, bc) = 1, as claimed.
(b) Suppose a = ck. Since (a, b) = 1 there exist integers x, y such that ax + by = 1. Thus c(kx) + by = 1. Since (c, b)|b and (c, b)|c, we have (c, b)|[c(kx) + by]. Thus, (c, b)|1, so (c, b) = 1.
Comments: You cannot use arguments involving prime factors in this problem because it is explicitly a Section 1 problem. Remember we discussed this in class: Do not use methods or concepts from later sections to solve problems in earlier sections — it defeats the whole purpose of giving you this work in Section 1 if you invoke prime factorization, which is not known until after that section. What we did here is clearly not the only approach to these problems but it makes nice use of a powerful result in the section.
23:S3:3 A says “We three have $100 altogether.” B says, “Yes, and if you had six times as much and I had one-third as much, we three would still have $100.” C says “It’s not fair. I have less than $30.” Who has what?
Solution: Let a, b, c be the amounts, in dollars, had by A, B, C respectively—presumably inte-gers. The three statements translate as
a + b + c = 100 6a + b
3+ c = 100 c < 30
The second statement is equivalent to 18a + b + 3c = 300. subtracting the first equation gives 17a + 2c = 200. An obvious particular solution: a0 = 0, c0 = 100.
Since (17, 2) = 1, the general solution is
a = 2t, c = 100 − 17t, t ∈ Z. Further,
b = 100 − a − c = 100 − 2t − (100 − 17t) = 15t.
Now our only constraints are that a, b, c ≥ 0 and c < 30. Thus, t ≥ 0 and 0 ≤ 100 − 17t < 30. That is, 7017 < t ≤ 10017. Since 7017 = 4 + 172 and 10017 = 5 +1517 it follows that t = 5. Therefore A, B, C have, respectively, a = 2 · 5 = $10, b = 15 · 5 = $75 and c = 100 − 17 · 5 = $15.
Comments: There are several possible variations. The first two equations could be subtracted to eliminate c, the denominator cleared by multiplication to obtain 15a − 2b = 0, and this can be solved parametrically. The matrix method may be used, but it is unwieldy simply because the particular solution obtained will involve multiplying by 300, and finding the right value for the parameter will involve unnecessarily large numbers.
