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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

its good to see you guys are having fun doing math. I hope this enthusiasm is infectious.

BTW, which is the 8th prob?

Ricky i dont understand 6. explain how u got m|n , m|n+1..i guess u have n|m²

another solution: write this as (m-2n)² = n(3n-1). gcd(n,3n-1) = 1. 3n-1 is a perfect square. contradiction! (source: AOPS)

The 8-th one is tricky!!

Prophet sir, hope u don't kill me if i say this thread was made by ricky and me for fun and to create some action......[3] For q (2) i spent sometime to see whether i can use IVT. Otherwise this is plain ...this was from ISI Bmath paper 2008.

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)

2) is quite trivial as you only need to prove that P(x) is a cubic

Proof: WLOG the leading coefficient of P(x) is positive. Let P(x) be of degree k

For large enough n, we have P(n)≥0 and so we can drop the mod.

Then the inequality becomes

cn^3 \le a_n n^k+a_{n-1}n^{k-1}+...+a_0 \le dn^3

or

c \le a_n n^{k-3}+a_{n-1}n^{k-4}+...+\frac{a_0}{n^3} \le d

Let us see what happens when n \rightarrow \infty

If k>3, we get d \ge \infty which is absurd

Again if k<3, we get c \le 0, but we are given c>0

Hence k=3.

But we know that any polynomial of odd degree has at least one real root and so we are done

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

6 . Write the given equation in the form -

Taking each side " modulo ' n ' " and " modulo ' m ' " , we arrive at -

It is easy to see that these set of conditions are never fulfilled by positive integers .

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)

''(since x1 * x2 =1 is not possible )'' why ?

x= ± 1√n-1

8)Find all function f:\mathbb{N}\to\mathbb{N} satisfying f(x+y)=f(xy) for all x,y\geq4 and f(5233) = 279

For Nishant bhaiya's hint

x₁ =x₂

or

x₁=1x₂

I think , the answer is x₁=x₂...=x_n=1√n-1
Not sure

5th has no soln in m,n