3b.) Use the division Algorithm.
Let x be any integer
⇒ if x divided by 9, then remainders are 0, 1, 2, 3, 4, 5, 6, 7, 8
⇒ 9k, 9k + 1, 9k + 2, 9k + 3, 9k + 4, 9k + 5, 9k + 6, 9k + 7, 9k + 8
To explain further: we have:
x = 9q : x3 = (3q)3
x3 = 27q 3
x3 = 9(3q 3 )
x3 = 9q, where q = 3q 3
x = 9q + 1 = x3 = (9q + 1)3
x3 = 729q 3 + 1
x3 = 9(3q 3 ) + 1
x3 = 9q + 1, where q = 3q 3 , and
x = 9q + 8 = x3 = (9q + 8)3
x3 = 729q 3 + 512
p
√
x3 = 729q 3 + 512
x3 = 9q + 8,
Since if satisfies the condition,
∴ The cube of any integers has one of the forms 9k + 1, or 9k + 8 ■
3c.) Let x be any integer
⇒ if x divided by 4, the remainders are 0, 1, 2, 3, 4
1
⇒ 5k, 5k + 1, 5k + 2, 5k + 3, 5k + 4
to prove further we have:
x = 5q : x4 = (5q)4
x4 = 625q 4
x4 = 5(125q 4 )
x4 = 5q, whereq = 125q 4 , and
x = 5q + 1 : x4 = (5q + 1)4
x4 = 625q 4 + 1
x4 = 5(125q 4 ) + 1
x4 = 5q + 1, whereq = 125q 4
Since it satisfies condition,
∴ The fourth power of any integers is either of the form 5k, or
5k + 1 ■
4.) Using contradiction, we have to assume that 3a2 −1 is a perfect square
Then 3a2 − 1 can be written as 3k or 3k + 1
⇒ 3a2 − 1 = 3k or 3a2 − 1 = 3k + 1
⇒ If 3a2 − 1 = 3k, then 3(a2 − k) = 1, since divide 1, then it is
contradiction.
⇒ If 3a2 − 1 = 3k + 1, then 3(a2 − k) = 2, since not divide 2,
then it is contradiction.
Since, it does not satisfies the form of 3k and 3k = 1
∴ 3a2 − 1 is never a perfect square. ■
2
5.) Mathematical Induction: 12 , 22 , 32 + .... + n2 =
n(n+1)(2n+1)
6
Proof:
For n=1, the statement reduces to 12
1×2×3
6
and is obviously
true.
Assuming the statement is true for n = k:
12 + 22 + 32 +...+(k + 1)2 =
(k+1)(k+2)(2k+3)
6
We will prove that the statement must be true for n = k+1: 12
+ 22 + 32 +....+ (k + 1)2 = (k+1)(k+2)(2k+3)
6
To simplify if, we have
(12 + 22 + 32 +...+(k + 1)2 =
(k+1)(k+2)(2k+3)
6
=
(k)(k+1)(2k+1)
6
=
(k)(k+1)(2k+1)+6(k+1)2
6
=
(k+1)[(k(2k+1)+6(k+1)]
6
=
(k+1)(2k2 +7k+6)
6
=
(k+1)(k+2)(2k+3)
6
+ (k + 1)2
Since the left-hand side is equal to the right-hand side
∴
n(n+1)(2n+1)
6
is an integer for every n ≥ 1. ■
6.) Using modular Arithmetic
Let n= 7m for some integer remainder −3 ≤ r ≤ 3;
We can have n=r (mod 7)
Since (−x)3 = (−x)3 , we have to check or finite
3
list of remainders:
03 = 0 (mod7)
13 = 1 (mod 7)
23 = 8 (mod 7)
33 = 27 (mod 7)
These clearly lie -1,0,1
∴ The cube of any integer is of the form of 7k or 7k ± 1 ■
8.) Proof:
Let x be an integer.
x=2n is the form of an even integer.
Square this even integer.
which is a multiple of 4. so, the number by the two right-most digit must
be divisible by 4.
Since, all given digits formed number 11, then it is not divisible by 4.
Let x be an odd integer.
x=2n+1 is the form of an odd integer.
square this odd integer.
Since, it is one more than a multiple of 4 then we have to subtract 1 from
number form of two-right most digits.
Performing this, we have 10 - which is not divisible by 4.
Since it is does not satisifies x=2n and x=2n+1
∴ no integer in the following sequence is a perfect square: II, III,
4
IIII, IIIII,... ■
10.) Proof by Mathematical Induction
Let n=1
1[7(12 ) + 5] = 7+5 = 12
Let n = m for some integer m ≥ 1
m(7m2 + 5) = 7m3 + 5m. ⇐= Let’s express as 6a.
We have to assume that 7m3 + 5m is divisible by 6.
Next, Let n= m+1
(m+1) (7(m + 1)2 + 5)
(m+1) (7m2 + 14m + 7 + 5)
(m+1) (7m2 + 14m + 12)
7m3 + 14m2 + 12m + 14m + 12
7m3 + 21m2 + 26m + 12
(7m3 + 5m) 21m2 + 21m + 12
6a +3 (7m2 + 7m4)
When m is odd, 7m2 + 7m + 4
When m is even, 7m2 + 7m + 4
Since, 7m2 + 7m + 4 is always divisible by 2
Then 3(7m2 + 7m + 4) is divisible by 6.
To explain further:
We let 3(7m2 + 7m + 4) be equal to 6b
5
= 6a+6b
= 6(a+b) ⇐= always divisible by 6
Since, it satisfies m and m+1, it’d be mean that it’ll work for 1,2,3,...+ z
∴ n (7n2 + 5) is the form of 6k for every n ≥ 1 (■
11.) Let n = 2k + 1, K ∈ N
Now, n4 + 4n2 + 11
= (2k + 1)2 + 4 (2k + 1)2 + 11
= (4k 2 + 4k + 1)2 + 4(4k 2 + 4k + 1) +11
= (16k 4 + 16k 2 + 1 + 32k 3 + 8k + 8k 2 ) + 16k 2 + 16k + 4 + 11
= 16k 4 + 32k 3 + 40k 2 + 24k + 16
= 16[k 4 + 2k 3 + 1]+ 40k 2 + 24k
Since,
= 16[k 4 + 2k 2 + 1]+
(80k2 +48k)
2
= 16[k 4 + 2k 2 + 1]+
16k(5k+3)
2
k(5k+3)
2
is integer for all N of K
so, we assume
k(5k+3)
=m
2
Now, n4 + 4n2 + 11
= 16[k 4 + 2k 2 + 1] +16m
= 16[k 4 + 2k 2 + 1 + m], where k=k 4 + 2k 2 + 1 + m
Since, 16 divides n4 + 4n2 + 11
∴ n4 + 4n2 + 11 is the form of 16k for every odd Z ■
4.) Assuming that gcd (a,b) = 1, Prove the ff:
6
a.) gcd (a+b, a-b) = 1 or 2
[Hint: let d = gcd (a + b, a − b) and shows that d|2a, d|2b, and thus that
d ≤ gcd (2a,2b) =2 gcd (a,b) Proof:
Let d|a + b, a-b, then d|(a + b) + (a − b)=2a, and
d|(a + b) − (a − b) = 2b so,
d|gcd (2a, 2b) = 2
Now, we assume that the greatest common divisor of a + b and a − b
is d.
This means, both a + b and a − b are divisible by d, i.e. d is a factor
of a + b and a − b.
Therefore, we can safely assume that a + b = p(d) and a − b = g(d).
Where p and g are other factors of a + b and a − b respectively.
Now, we will add p(d) and g(d)
=⇒ 2a = p(d) + g(d)
=⇒ 2a = d(p + q)
From the above equation, We can say that d is a factor of 2q.
Similarly, We will now subtract g(d) from p(d)
=⇒ 2b = p(d) − g(d)
=⇒ 2b = d(p − g)
Thus, We can also say that d is a factor od 2b.
Therefore, this is only posible if either d = 1or2 Hence, GCD (a + b, a − b)
= 1 or 2. ■
b.) gcd (2a + b, a + 2b) = 1 or 3
7
Since, gcd (a, b) = 1, There exists integer x, y such that ax + by = 1
Proof: Suppose gcd (a,b) = 1
Let d = gcd (2a + b, a + 2b).
Then d|(2a + b) and d|(a + 2b).
d|2(2a + b) - (a + 2b) =⇒ d|3a
d|2(a + 2b) - (2a + b) =⇒ d|3b
then,
d|3a =⇒ d|3ax
d|3b =⇒ d|3by
Hence,
d|(3ax + 3by)
=⇒ d|[3(ax + by)]
=⇒ d|3[sinceax + by = 1]
=⇒ d = 1 or d = 3.■
c.) gcd (a + b, a2 + b2 ) = 1 or 2
Proof:
Suppose gcd (a, b) =1
Let d = (a + b, a2 + b2 )
I f d divides a + b then it divides a(a + b) = a2 + ab so if it also divides
a + b2 , then it divides,
2
a2 + ab - (a2 + b2 ) = ab − b2 =b(a − b)
8
A similar calculation shows it divides a(a − b)
Then from gcd (a, b) = 1, we get d divides a − b:
then it divides
(a + b) + (a − b) =2a and
(a + b) + (a − b) = 2b, so
d = 1 or d = 2 ■
d.) gcd (a +b, a2 − ab + b2 ) = 1or 3
Proof:
Let d = a2 − ab + b2
Since d|(a + b) and d|(a2 − ab + b2 )
=⇒ d|[(a + b)2 − a2 − ab + b2 ] =⇒ d|3ab
Therefore, d|[3b(a + b) − 3b], i.e, d|3b2
Similarly it can be shown that d|3a2
Therefore, d|(3a2 , 3b2 ) =⇒ d|3(a, b)2 =3
Hence, (a + b, a2 − ab + b2 ) = 1 or 3. ■
5.) For n ≥ 1, and positive integers a,b,
Show the following:
a.) If gcd (a, b) = 1, then gcd (an , bn ) = 1
Proof:
Suppose gcd (a, b) = 1 =⇒ gcd (a2 , b2 ) = 1
gcd (a,b) = 1 implies
9
ax + by = 1
a2 x2 = 1 − 2by + b2 y 2
(2y − by 2 )b = 1 − a2 x2
(2y − by 2 )2 b2 = 1 − a2 x2 + a4 x4
(2x2 − a2 x4 ) a2 + (2y − by 2 )2 b2
∴ gcd (a2 , b2 ) =1 ■
b.) The relation an |bn implies that a|b.
Proof by Mathematical Induction: Lets denote the set of all positive integers n for which an |bn =⇒ a|b.
We observe that when n=1 , a|b =⇒ a|b.
This mean that 1 ∈ S.
Next, we suppose that K ∈ S so that ak |bk .
We need to show that ak+1 |bk+1 =⇒ a|b.
i.e, that k + 1 ∈ S.,.....
To show this, we assume that ak+1 |bk+1 .
Which is precisely the case for n = k + 1 putting k + 1 ∈ S whenever
K ∈ S.
According to the first principle of finite induction; S must be the set of all
positive integers. ■
9.) Prove that the greatest common divisor of two positive integers divides
their least common multiple.
Proof:
Let gcd (a, b)=d, lcm (a, b) = m
10
Since d|a and d|b, also, a|m and b|m
So, d|m, ( by transitivity of division)
=⇒ a=nd or b=md for some integers m and n
gcd (a, b) × lcm(a, b) = ab
ab
=
=⇒ lcm (a,b) = gcd(a,b)
(nd)(md)
d
= d(mn)
=⇒ d|lcm(a, b) ∴ mn ∈ Z
=⇒ gcd (a, b) | lcm(a, b) ■
12. Find integers x, y, z Satisfying gcd (198, 288, 512) = 198x+288y +512z
Proof:
Let d = gcd(198, 288)
288 = 98 × 1 + 90
198 = 90 × 2 + 18
90 = 18 × 5 + 0
So, gcd (198, 288) = 18
18 = 198 − 2 × 90
=198 − 2(288 − 198)
= 3 × 198 − 2 × 288
Now, gcd (198, 288, 512) = gcd (918, 512)
512 = 18 × 28 + 8
18 = 2 × 8 + 2
8 = 4 ×2 = 04
So, gcd (18, 512)=2
2 = 18 − 2 × 8
11
= 18 − 2[512 − (8)(28)]
=57 × 18 − 2 × 512
= 57(3 × 198 − 2 × 288) −2 × 512
= 171 × 198 − 114 × 288 − 2 × 512
x = 171, y = −114, z = −2 ■
12