.. it comes from the Fourier series theory .

 As usual, assume for positive integers m,n the lengths of the side of the right angled triangles are (m^2-n^2), 2mn, (m^2+n^2) , with (m^2+n^2) being the hypotenuse. We have to prove that for all integers m,n (m>n)

area of the triangle S(m,n)=mn(m+n)(m-n) is divisible by 6.

First we prove that S(m,n) is divisible by 2

if either m or n is even then it holds naturally

else

if both m and n are odd then both (m+n) and (m-n) are even

Hence S(m,n) is divisible by 2.                       .... (1)

Next we show that S(m,n) is is divisible by 3  

if either m or n is divisible by 3 then it holds naturally.

else if m= 3k + 2 and n= 3l + 1

or, m = 3k+ 1 and n = 3l +2 (for some integers k,l) then m+n is divisible by 3

on the other hand if m= 3k+1 and n =3l +1 or,

m= 3k+2 and n= 3l+2  (for some integers k,l) then m-n is divisible by 3

this exhausts all the possibilities

Hence S(m,n) is divisible by 3 also           .... (2)

Combining (1) and (2) we conclude that the area of the triangle is divisible by 6. (Proved)

If possible, suppose that f(x) has an integral root , let it be "b"

then we may write

      f(x)=(x-b)*Q(x) (where Q(x) is a polynomial of degree one less than f(x) )

ow since f(x) is a polynomial with integer coeffiecients, it follows (by just multiplying out term by term, Q(x) is also a polynomial with integral coeffiecients)          ..... (1)

So, we have

f(0)*f(1)=(-b)*(1-b)*Q(0)*Q(1) = odd number (since f(0) and f(1) are both odd)

also Q(0) and Q(1) are integers (by (1)) and (-b) and (1-b) are two consecutive integers, hence there product is always even.

Hence (-b)(1-b)*Q(0)*Q(1) is an even number .

Hence contradiction.

apply Taylor series expansion for sin x, sin 3x etc upto 5th degree and take limit as x tends to zero .... it will give you condition on the existance of limits involving A and B

Clarification ....

I think ppl are confused about the problem statement ....

It is f(x)=f(2-x)

and NOT f(x)=f(2-x)*f(x)

hope this helps.

Ans : 6

From the relation , it is clear that if b is a root then 2-b is also a root. Sum of these pair of roots = 2.

There are 3 such pairs. Hence sum of the roots =6

apply modulo arithmatic ..

4^2=1mod(3)

raising both sides to the power of 43, we get

4^86=1mod(3)

so 4^87=4mod(3)=1mod(3)

Hence the remainder is 1

 because of radiation resistance 

 mean square value of current = 1/tao int (0 to tao) io^2 t^2/tao^2 dt = io^2/3

so rms= io/root(3)

 Wrong statement ......

All the four equations of Maxwellian Electrodynamics can be derived starting from Coulomb's law and then applying relativistic transformations.

What's the use of two diodes in the last circuit ??

 obviously possible.

the only requirement is that its curl does not identically vanish everywhere.

e.g. E(r) = -yi + xj  is an example of a  non conservative field.

simple .. only (C) is correct.

we have @1 +@2 + .. +@8 =0 ... ( (A) can't be correct )

(@1)^3 + (@2)^3 + .... +(@8)^3 = 3(@1 . @2 .... @8) =3 ( (B) can't be correct )

also see that (@1)^8 = (@2)^8 = .... (@8)^8=1

so (@1)^48 + (@2)^48 + ... (@8)^48 = 8 ( (D) can't be correct )

Now observe that each of (@i)^2 is a 4th root of unity .

So (@1)^50 + (@2)^50 +.... (@8)^50

= (@1)^2 + (@2)^2 + ... (@8)^50

= 2* (sum of the 4th roots of unity)

=0

So (C) is the ONLY correct answer.

pi = 4 * int (0 to 1) dx/(1+x^2)  

Now expand the binomial in the Denominator and integrate term by term

(which is valid as the series converges within the limits mentioned)

This is the famous multinomial theorem ... search for it in Wiki

Apply Cardan's rule

Evaluate :

Where C(n,m) denotes the usual combinational term nC_m

The cases are mutually exclusive and exhaustive and both are proved means the theorem is proved . It is another style of proving a lemma or theorem and sometimes the only feasible way to the solution .

@ Hari Shankar

The question says only positive integers

@Rahul

a very good answer indeed.

But the result can be obtained with much less effort and in a more general way

case 1. a=b

in this case u get a =b = sqrt (3) ; hence no soln

case 2. a!=b

sice the expression is symmetric wrt a and b , without any loss of generality , we may assume a > b

then we get

 1 < 3/b^2

or b< sqrt(3)

the only possibility is then b = 1

which gives 1/a^2 + 1/a = 0 , which gives no solution for positive integer 'a' as the LHS > 0

(proved)