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

2011/010) Find integral solution of (1-i)^n = 2^n

We have 1-i = (sqrt(2) cis (-pi/4)

So (1-i)^n = 2^(n/2) cis (-npi/4) = 2^n

The modulo of LHS = 2^(n/2) and rhs = 2^n and both are same

If 2^(n/2) = 2^n or n = 0

Then = 2^(n/2) cis (-npi/4) = 1 = RHS

So n = 0 is the only solution

2011/009) one problem in AP

question

The fourth power of the common difference of an AP with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

proof

With out loss of generality we can take the 1st term of the 4 consecutive terms to be a and let the difference be t

a and t are integers

We have the 4 terms = a, a+t, a+2t, a+ 3t and 4th power of difference is t^4

Now a(a+t)(a+2t)(a+3t) + t^4

= (a(a+3t))((a+t)(a+2t)) + t^4
= (a^2+3ta)(a^3+3ta + 2t^2) + t^4

Letting a^2 + 3ta = p we get

= p(p+2 t^2) + t^ 4= (p^2+ 2pt^2+t^4) = (p+t^2)^2 = (a^2+3at+t^2)^2

Which is square of integer as a^2+3at+t^2 is integer