1. ## Mathematics

P^(n+1) = 2011 + Q^(n+1)

find all positive integer sets (p,q,r)

how I do this all day and i can't even fuck start

2. All (P, Q, n), maybe? (I don't see an r.)

There is an identity: for any numbers x and y and any positive integer m > 2,

x^m - y^m = (x - y)(x^(m-1) + x^(m-2) y + x^(m-3) y^2 + ... + x y^(m-2) + y^(m-1))

If x and y are integers, then the factor after (x - y) is also an integer. So in this case, for any m > 1, x^m - y^m is an integer multiple of x - y.

So, for any positive integer n, and any positive integers P and Q, P^(n+1) - Q^(n+1) is an integer multiple of P - Q. As P^(n+1) = 2011 + Q^(n+1) we have P^(n+1) > Q^(n + 1) and hence P > Q. So P - Q is a _positive_ integer that divides 2011. Now 2011 only has two positive integer factors: 2011 and 1.

If P - Q = 1, let's stare again at

2011 = P^(n+1) - Q^(n+1) = (P - Q)(P^n + P^(n-1) Q + ... + P Q^(n-1) + Q^n).

as P - Q = 2011, the sum P^n + P^(n-1) Q + ... + P Q^(n-1) + Q^n has to be 1. But it's sum of n+1 positive integers. The only way it can be 1 is if n = 0, and we were told only to consider positive n, so we have no solutions with P - Q = 2011.

Let's find the solutions with P - Q = 1.

First let's see what we find if n + 1 is even. In this case, there is a positive integer k with n + 1 = 2k. From the formula for a difference of squares we have

2011 = P^(2k) - Q^(2k) = (P^k - Q^k)(P^k + Q^k),

so P^k + Q^k is also a divisor of 2011. As it is at least 2, it must be 2011, so P^k - Q^k has to be 1. If k is bigger than 1, applying the algebraic identity above with x = P and y = Q and m = k and using the fact that P - Q = 1 we get

1 = P^k - Q^k = P^(k-1) + P^(k-2) Q + ... + Q^(k-1)

but the sum on the right, being a sum of k > 1 positive integers, cannot be 1. So in fact k = 1, and n + 1 = 2, so the equation reads 2011 = P^2 - Q^2, which since P - Q = 1 is 2011 = (Q + 1)^2 - Q^2 = 2Q + 1. Solving for Q we find Q = 2010/2 = 1005 and thus P = 1006. So (P,Q,n) = (1006,1005,1) is one solution, and it is the only solution with P - Q = 1 and n+1 even.

If P - Q = 1 and n+1 is odd, let's look again at

2011 = (Q + 1)^(n+1) - Q^(n+1)

If you think of expanding the right hand side it has the form 1 + [a multiple of Q], so 2011 is 1 + a multiple of Q, so 2010 is a multiple of Q. If you rewrite the right hand side as P^(n+1) - (P - 1)^(n+1), it takes the form [a multiple of P] - (-1)^(n+1) which is [a multiple of P] - (-1) = [a multiple of P] + 1 since n+1 is odd. So 2010 is also a multiple of P.

So only need to check among the pairs (P,Q) with P and Q positive divisors of 2010 and P - Q = 1.

2010 = 2*3*5*67, so 2010 has only 16 positive divisors:

1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010

Only three pairs (P,Q) have the property that P - Q = 1, (2,1), (3,2), and (6,5).

In the case P = 2, Q = 1 we are solving 2011 = 2^(n+1) - 1, which is easily seen to have no solutions since 2011 + 1 = 2012 is not a power of 2.

In the case P = 3, Q = 2 we are solving 2011 = 3^(n+1) - 2^(n+1), and we can check for n <= 5 the right hand side is smaller than 2011, and for n > 5 the right hand side is bigger than 2011.

The case P = 6 and Q = 5 is similar: for n <= 3 you can see that 6^(n+1) - 5^(n+1) is too small, and for n > 3 it is too big.

So there are no solutions with P - Q = 1 and n+1 odd.

So there's only one solution to this in the positive integers, (P, Q, n) = (1006, 1005, 1).

The way I've solved it is by no means the only way to solve it; it's just what came to me as I was working on the problem.

3. ## The Following User Says Thank You to Registration For This Useful Post:

[MPGH]Wyo (12-01-2011)

4. such a pro dude Thanks.