2011/020) Gaussian Integer

Here i discuss Gaussian integer and property of Gaussian integer

x+iy in complex number is a Gaussian integer if x and y are integers

they are closed under addition subtraction and multiplication

A Gaussian integer a + bi is prime if and only if:

* one of a, b is zero and the other is a prime of the form 4n + 3 or its negative − (4n + 3) (where n \geq 0)
* or both are nonzero and a2 + b2 is prime.


13 is not a Gaussian prime because 13 = 9+4 = (3^2+2^2) = (3+2i)(3-2i)

any prime number of the form 4n+1 can be expressed as sum of 2 squares and cannot be a Gaussian prime.

but a prime of the form 4n+3 cannot be expressed as sum of 2 squares and hence Gaussian prime.

2 is a special prime as it is not odd and 2 = 1^2+1^2 = (1+i)(1-i)

2011/019) Prove that Sin (a +b ) sin (a-b) = sin ^2 a – sin ^2b

The above is an interesting identify in trigonometry and this can be done wrongly with out understanding trigonometry

As we do not know trigonometry we take sin as a multiplier

So sin (a+b) = sin a + sin b

And sin (a-b) = sin a – sin b

Multiply them to get

Sin (a+b) sin (a-b) = (sin a + sin b) (sin a - sin b)
= sin ^2 a – sin ^2 b

however above is wrong but the ans is right

now to the correct approach

Sin (a+b) sin (a-b)
= (sin a cos b + cos a sin b)(sin a cos b – cos a sin b)
= sin ^2 a cos^2b – cos^2 a sin ^2b
= sin ^2 a(1- sin ^2 b) – cos^2 a sin ^2 b
= sin ^2 a – sin ^2 a sin ^2 b – cos^2 a sin^2 b
= sin ^2 a – sin ^2 b(sin ^2 a + cos^2 a)
= sin ^2 a – sin ^2 b


The above example is to demonstrate that we get right result through erroneous approach and one need to be careful about it

2011/018) Prove that in the product

(1-x+x^2-x^3+ ..... - x^99 + x^100) ( 1 +x +x^2 + .. x^99 + x^100) after multiplying there does not appear a term of x odd degree.

this problem is solved by taking coefficient in a couple of books and I try to solve in a different approach

LHS = (1-x^101)/(1-x) * (1+ x^101)/(1+x)
= (1-x^202)/(1-x^2)

if we define f(x) = (1-x^101)/(1-x) then f(x^2) is the given expression and as it is division of p(x^2) by q(x^2) so cannot have any term other than (x^2)^2k or odd power of x

2011/017) if ( p and q) are roots of equation (x-a)(x-b) = c then

-->

Roots of equation (x-p)(x-q) + c are

a) a c

b) bc

c) a,b

d) (a+c), (b+c)

solution

-->

we have

(x-a)(x-b)) - c = x^2- (a+b) x+ (ab-c)

As p and q are roots so p+ q = a + b …1

and pq = ab – c or pq + c = ab …2

Now

(x-p)(x-q) + c = x^2 - (p+ q) x + (pq + c)

= x^2-(a+b)x + ab = (x-a)(x-b) ( from 1 and 2)

So roots are a and b

hence ans is c)

2011/016) a problem with repunit

A number is called repuint if all the digits of the number are 1 that is it is 1 , 11 , 111 so on

in a repunit number with 50 1's the 26th digit from the left is modified to some digit so that is is divisible by 13. what is the digit

Ans:
method 1
say the digit x
the number is 1(25 times)x1(24 times)
= (10^50-1)/9 + (x-1)(10^25-1)/9

we need to find x such that
(10^50-1)/9 + (x-1)(10^25-1)/9 mod 13 = 0

as 9 is coprime to 13

so (10^50-1)+ (x-1) )(10^25 -1)mod 13 = 0


as 10 and 13 are coprime so as per format's little theorem


10^12 = 1 mod 13

so 10^48 = 1 mod 13

so 10^50 = 100 mod 13

so (10^50-1)+ (x-1) )(10 -1) =

(100-1) + 9(x-1) mod 13 = 0

99 + 9x - 9 or 90+9x = 0 mod 13 or 10 +x = 0 mod 13 or x= 3

so ans = 3

Method 2


sequence of 12 1's is divisible by 13 so we subtract sequence of 12 1's and divide by 10^12 and do it 2 times

we get 1...(25times)x

as seqeunece of sequence of 12 1's is divisible by 13 so multiplying by any number so multiply by 10^14 and subtract

we get 1..(13 times) x

as sequence of 12 1's is divisible by 13 so multiplying by any number so multiply by 10^2 and subtract

1x is the number and divisible by 13 so x = 3

2011/015) Let p be a prime number > 5. Show that p divides infinitely many of the numbers 1, 11, 111 ...

as p > 5 p and 10 are coprimes and so are p and 9

so if p devides x <=> p devides 9x

so p devides 9, 99, 999, so on

so 10^n mod p = 1

now as p and 10 are co-primes then

10^n(p-1) is 1 mod p(for n = 1 ... )

or 10^n(p-1) -1 is divsible by p

or (10^n(p-1)-1)/9 that is 1 repeated n(p -1) is dvisible by p

so there are many numbers divisible by p

2011/014) Proof of existence of infinite prime numbers.

There are infininite number of prime numbers and there are a couple of proofs these are in the linked list below

http://primes.utm.edu/notes/proofs/infinite/index.html

I propose a proof that is simpler

If I prove that if for any n there is a prime > n then I am through

Let us consider n!+1

This does not have any factor from 2 to n and hence it is a prime or in case it is not a prime then prime factor > n.

So for any n there is a prime number> n

Hence proved

2011/013) factorization using short cuts.

1) (a+b+c)^3 - a^3-b^3-c^3

we know that (a+b+c)^3 - a^3 is divisible by (a+b+c) – a that is b+ c

and b^3+c^3 by (b+c)

hence b+c is a factor

by symmetry (a+b) and (c+a) are also factors

2) so it is (a+b+c)^3 - a^3-b^3-c^3 = m(a+b)(b+c)(c+a)

now as LHS does not contain a^3 or b^3 or c^3 (as they cancel out) so m has to be a constant

putting a =1 b = 1 and c = 1 we get LHS = 24 and RHS = 8m or m = 3

hence (a+b+c)^3 - a^3-b^3-c^3 = 3(a+b)(b+c)(c+a)

2011/012) To construct an AP of integers so that 3 successive elements are perfect squares.

It is proved that we cannot have an AP whose 4 successive terms are in perfect squares
But does there exist an AP whose 3 consecutive terms are perfect squares

Solution:

Let the 3 consecutive term be a^2,b^2,c^2

As they are in AP we have
b^2-a^2 = c^2-b^2
or a^2+c^2 = 2b^2

this has a solution and we know that

if x^2 + y^2 = z^2

then (x+y)^2 + (x-y)^2 = 2(x^2+y^2) = 2z^2

so if (x,y,z) is a Pythagorean triplet the (x-y)^2 , z^2, (x+y)^2 are perfect squares and are in AP.

Or a= x-y
b = z^2
v= x+ y
for example
(3,4,5) is Pythagorean triplet so (4-3)^2, 5^2,(4+3)^2 or 1,25,49
(5,12,13) Pythagorean triplet so (12-5)^2, 13^2,(12+5)^2 or 49,169,289

Parametric form of Pythagorean triplet is
(m^2-n^2), (2mn), m^2 + n^2

So Parametric form of the required AP is

(m^2-n^2-2mn)^2,(m^2+n^2)^2,(m^2-n^2+2mn)^2

2011/011) what is the smallest prime factor of 2010! +1

As per Wilson's theorem for prime p

(p- 1)! = - 1 mod p

as as 2011 is a prime number

2010 ! = - 1 mod 2011

so 2011 is a factor of 2010 ! + 1

and as 2 to 2010 cannot divide it so smallest prime factor is 2011