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3 .

solution to problem 8. let x≥4 ,f(4+x) = f(4x) = f(2x+2x)= = f(4x²) = f(4x.x) = f(4x+x) = f(5x)= f(5+x)[4]

prob 8=# 9 (btw nokia '5233' is a phone that i had,its a prime)

infectious! :P

When i looked over the solution again i realized that you dont need that

x≡2 mod p

One of the factors divides the other is all that you need.

x^5+x^4+1 = (x^2+x+1)(x^3-x+1)=p^y

Now, we clear up the case when one of the factors equals 1.

For natural numbers x, we have x^2+x+1 \ge 3 so we only need to check the other factor and we see that equality occurs when x=1.

Verifying, we see that we have (1,1,3) as one solution.

Now, if both the products are greater than 1, we must have

x^2+x+1 = p^{n_1}; x^3-x+1 = p^{n_2} with n_1+n_2=y

So p divides both numbers so that it also divides

(x-1)(x^2+x+1)-(x^3-x+1)=x-2

i.e. x \equiv 2 \pmod p

Now,

x^2+x+1 \equiv 2^2+2+1 \equiv 7 \equiv 0 \pmod p \Rightarrow p=7

So, we have

x^2+x+1 = 7^{n_1}; x^3-x+1=7^{n_2}

That means x^2+x+1 |x^3-x+1 \Rightarrow x^2+x+1|x-2

Since x^2+x+1 > x-2 this is possible iff x=2

Verifying we see that (2,2,7) is also a solution

Thus the only solution sets are (1,1,3) and (2,2,7)

7)

First note that m and m-2 are relatively prime as otherwise their gcd would divide m - (m-2) = 2, which would mean gcd =2 which is absurd as m is odd.

So it suffices to separately prove divisibility by m and m-2

Now by Euler Totient theorem we have

2^{\phi(m)} \equiv 1 \pmod m

Since\phi(m) \le m-1, we have \phi(m)| (m-1)!

Hence it follows that 2^{(m-1)!} \equiv 1 \pmod m

By a similar argument 2^{(m-1)!} \equiv 1 \pmod {(m-2)}

m and m-2 being relatively prime we have

2^{(m-1)!} \equiv 1 \pmod {m(m-2)}

thus proving the assertion

krishna..i typed the q wrong..i hv edited it ...

since if x1 *x2 =1 , then comparing similar results we have x1=x2 =....... = x_n =±1
which does not satisfy the reqd. condition

and of course x will have ± 1/√(n-1)

adding all lhs and rhs ,
(n-1)\sum{x_{i}} = \sum{\frac{1}{x_{i}}} ... (i)

using eq. (i) in the first eqn ,
(n-1)(x_{1} +1/x_{1}) = \sum{\frac{1}{x_{i}}}
similarly ,
(n-1)(x_{2} +1/x_{2}) = \sum{\frac{1}{x_{i}}}

similarly for others,

dividing these eqns ,
we have ,
(x_{2} +1/x_{2}) = (x_{1} +1/x_{1})

or ,x_{1} =x_{2} (since x₁ * x₂ =1 is not possible )

hence, = x (say)

replacing in any eqn ,
(n-1).x = 1/x

or, x = \frac{1}{\sqrt{n-1}}

A try
Let us consider a case

x₁=x₂=x
x₁=x₃ =x
.
.
.
x₁=1x_i ≡x_i=1x
.
.
.

From one of the conditions

kx+lx=0

no real solutions for x

??

so only possible solution is

x₁=x₂=x₃=x₄.....x_n=1√n-1

is it right ?

yup..but u should post solution...:D