Chapter
3
Arithmetic
3.1 Division Algorithm
169 Definition If a
6= 0,b are integers, we say that a divides b if there is an integer c such that ac = b. We write this as a|b. If a does not divide b we write a
6 |b. It should be clear that if a|b and b 6= 0 then 1 ≤ |a| ≤ |b|.
170 Theorem The following are properties of divisibility.
• If c divides a and b then c divides any linear combination of a and b. That is, if a,b,c,m,n are integers with c|a,c|b, then c
|am + nb. • Division by an integer is transitive. That is, if x,y,z are integers with x|y,y|z then x|z.
Proof: There are integers s ,t with sc = a,tc = b. Thus
am + nb = csm + tn ,
giving c |am + bn. Also, there are integers u,v with xu = y,yv = z. Hence xuv = z, giving x|z. ❑
A very useful property of the integers is the following:
171 Theorem Division Algorithm Let a
, b be integers, b 0. There exist unique integers q and r satisfying a = bq + r
, 0 ≤ r b 3.1
Proof: The set S = {a − bs : s ∈ Z,b − as ≥ 0} is non-empty, since a − b−a
2
≥ 0. Since S is a non-empty set of non-negative integers, it must contain a least element, say r = a − bq
. To prove uniqueness, assume a = bq + r = bq
′
+ r
′
with 0 ≤ r
′
b. Then bq − q
′
= r
′
− r. This means that b |r
′
− r. Since 0 ≤ |r
′
− r | b, we must have
r
′
= r. But this also implies q = q
′
. ❑ For example, 39 = 4
· 9 + 3. The Division Algorithm thus discriminates integers according to the remainder they leave upon division by a. For example, if a = 2, then according to the Division Algorithm, the integers may be decomposed into the two
families A
= {... − 4,−2,0,2,4,...},
A
1
= {...,−5,−3,−1,1,3,5,...}.
Therefore, all integers have one of the forms 2k or 2k + 1 . We mention in passing that every integer of the form 2k + 1 is
also of the form 2t − 1 , for 2k + 1 = 2k + 1 − 1, so it suffices to take t = k + 1.
24
Division Algorithm 25
If a = 4 we may decompose the integers into the four families B
= {... ,−8,−4,0,4,8,...},
B
1
= {... ,−7,−3,1,5,9,...},
B
2
= {...,−6,−2,2,6,10,...},
B
3
= {...,−5,−1,3,7,11,...}.
Therefore any integer will take one of the forms 4k , 4k + 1, 4k + 2 or 4k + 3. Again, any integer of the form 4k + 1 is also of the
form 4t − 3 and any integer of the form 4k + 3 is also of the form 4t − 1.
172 Example Shew that the square of any integer is of the form 4k or of the form 4k + 1. That is, the square of any integer is
either divisible by 4 or leaves remainder 1 upon division by 4. Solution: If n is even, that is n = 2a
, then n
2
= 2a
2
= 4a
2
, which is of the form 4k. If n is odd, say n = 2t + 1, then n
2
= 2t + 1
2
= 4t
2
+ t + 1 , which is of the form 4k + 1.
173 Example Shew that no integer in the sequence
11 , 111, 1111, 11111, . . .
is a perfect square. Solution: Clearly 11 is not a square, so assume, that an integer of this sequence has n
2 digits. If n 2, 11
. . . 1
| {z
}
n 1
′
s
= 11 . . . 11
| {z
}
n−2 1
′
s
00 + 12 − 1 = 100 · 11...11
| {z
}
n−2 1
′
s
+12 − 1 .
Hence any integer in this sequence is of the form 4k − 1. By the preceding problem, no integer of the form 4k − 1 can be a square. This finishes the proof.
174 Example Shew that n
2
+ 23 is divisible by 24 for infinitely many values of n. Solution: Observe that n
2
+ 23 = n
2
− 1 + 24 = n − 1n + 1 + 24. Therefore the families of integers n = 24m ± 1,m =
, ±1,±2,±3,... produce infinitely many values such that n
2
+ 23 is divisible by 24.
175 Example
Shew that the square of any prime greater than 3 leaves remainder 1 upon division by 12 .
Solution: If p 3 is prime, then p is of one of the forms 6k ± 1.
Now, 6k
± 1
2
= 123k
2
± k + 1, proving the assertion.
176 Example Prove that if p is a prime, then one of 8p − 1 and 8p + 1 is a prime and the other is composite.
Solution: If p = 3 , 8p −1 = 23 and 8p +1 = 25, then the assertion is true for p = 3. If p 3, then either p = 3k +1 or p = 3k +2.
If p = 3k + 1 , 8p − 1 = 24k − 7 and 8p + 1 = 24k − 6, which is divisible by 6 and hence not prime. If p = 3k + 2, 8p − 1 =
24k − 15 is not a prime, .
177 Example
Shew that if 3n + 1 is a square, then n + 1 is the sum of three squares.
26 Chapter 3
Solution: Clearly 3n + 1 is not a multiple of 3, and so 3n + 1 = 3k ± 1
2
. Therefore n + 1 =
3k ± 1
2
− 1 3
+ 1 = 3k
2
± 2k + 1 = k
2
+ k
2
+ k ± 1
2
, as we wanted to shew.
178 Example AHSME 1976
Let r be the common remainder when 1059 , 1417 and 2312 are divided by d 1. Find d − r.
Solution: By the division algorithm there are integers q
1
, q
2
, q
3
with 1059 = dq
1
+ r , 1417 = dq
2
+ r and 2312 = dq
3
+ r. Subtracting we get 1253 = dq
3
− q
1
, 895 = dq
3
− q
2
and 358 = dq
2
− q
1
. Notice that d is a common divisor of 1253 , 895,
and 358. As 1253 = 7 · 179, 895 = 5 · 179, and 358 = 2 · 179, we see that 179 is the common divisor greater than 1 of all three
quantities, and so d = 179 . Since 1059 = 179q
1
+ r , and 1059 = 5 · 179 + 164, we deduce that r = 164. Finally, d − r = 15.
179 Example
Shew that from any three integers, one can always choose two so that a
3
b − ab
3
is divisible by 10. Solution: It is clear that a
3
b − ab
3
= aba − ba + b is always even, no matter which integers are substituted. If one of the three integers is of the form 5k, then we are done. If not, we are choosing three integers that lie in the residue classes 5k
± 1 or 5k
± 2. By the Pigeonhole Principle, two of them must lie in one of these two groups, and so there must be two whose sum or whose difference is divisible by 5. The assertion follows.
Practice
180 Problem Find all positive integers n for which
n + 1 |n
2
+ 1 .
181 Problem If 7
|3x + 2 prove that 7|15x
2
− 11x − 14 ..
182 Problem Shew that the square of any integer is of the form 3k or 3k + 1
.
183 Problem Prove that if 3
|a
2
+ b
2
, then 3 |a and 3|b
184 Problem Shew that if the sides of a right triangle are all integers, then 3 divides one
of the lengths of a side.
185 Problem Given that 5 divides n + 2, which of the following are divisible by 5
n
2
− 4 , n
2
+ 8n + 7 , n
4
− 1 , n
2
− 2n?
186 Problem Prove that there is no prime triplet of the form p
, p + 2, p + 4, except for 3
, 5, 7.
187 Problem Find the largest positive integer n such that
n + 1n
4
+ 2n + 3n
3
+ 57 be divisible by n
2
+ 2 .
188 Problem Demonstrate that if n is a positive integer such that 2n + 1 is a square,
then n + 1 is the sum of two consecutive squares.
189 Problem Shew that the product of two integers of the form 4n + 1 is again of this
form. Use this fact and an argument by contradiction similar to Euclid’s to prove that there are infinitely many primes of the form 4n − 1.
190 Problem Prove that there are infinitely many primes of the form 6n − 1
.
191 Problem Prove that there are infinitely many primes p such that p − 2 is not prime.
192 Problem
Demonstrate that there are no three consecutive odd integers such that each is the sum of two squares greater than zero.
193 Problem Let n
1 be a positive integer. Prove that if one of the numbers
2
n
− 1 , 2
n
+ 1 is prime, then the other is composite.
194 Problem Prove that there are infinitely many integers n such that 4n
2
+1 is divisible by both 13 and 5.
195 Problem Prove that any integer n
11 is the sum of two positive composite num- bers.
196 Problem Prove that 3 never divides n
2
+ 1 .
197 Problem
Shew the existence of infinitely many natural numbers x , y such that
xx + 1 |yy + 1 but
x 6 |y and x + 1 6 |y,
and also x
6 |y + 1 and x + 1 6 |y + 1.
The Decimal Scale 27
Kategori