{P-Q}{P^(n-1) + K1[P^(n-2)]Q + K2[P^(n-3)]Q^2 + ....+ Q^(n-1)} cannot equal C^n - B^n

where all the variables are positive integers, n>2, P>Q, C>B

unless all the Ks are equal to 1 or in some instances the Ks have the same value.

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- Jul 23rd 2013, 07:37 AMMrAwojobiProof of a Conjecture or Counter Examples
{P-Q}{P^(n-1) + K1[P^(n-2)]Q + K2[P^(n-3)]Q^2 + ....+ Q^(n-1)} cannot equal C^n - B^n

where all the variables are positive integers, n>2, P>Q, C>B

unless all the Ks are equal to 1 or in some instances the Ks have the same value.

- Jul 23rd 2013, 09:02 AMMrAwojobiRe: Proof of a Conjecture or Counter Examples
Can anyone please prove or disprove this?

- Jul 23rd 2013, 11:19 AMMrAwojobiRe: Proof of a Conjecture or Counter Examples
Please can I have another counterexample, if any, because this one kind off eliminates Q by setting it equal to 1. So an additional constraint of Q > 1.

- Jul 23rd 2013, 11:30 AMtopsquarkRe: Proof of a Conjecture or Counter Examples
Please don't bump your threads.

-Dan - Jul 23rd 2013, 11:33 AMChessTalRe: Proof of a Conjecture or Counter Examples
Can i ask why my post was deleted?(Worried) It was a completely legitimate post providing a proper counterexample.

PS: I have another counterexample with the new constraint given, but if someone is deleting links for no reason, there is no reason to post it.(Headbang)

EDIT:

Forget my above comments. It is just the TS created 2 exact topics and i answered to the other:

http://mathhelpforum.com/number-theo...-examples.html - Jul 23rd 2013, 11:45 AMMrAwojobiRe: Proof of a Conjecture or Counter Examples
So what's your other counterexample please?

- Jul 23rd 2013, 11:51 AMtopsquarkRe: Proof of a Conjecture or Counter Examples
First, take a look at the example in the other thread and see if you can come up with another one based on ChessTal's method.

This is why we don't double post....

-Dan - Jul 23rd 2013, 11:54 AMChessTalRe: Proof of a Conjecture or Counter Examples
For:

P=3

Q=2

n=3

K1=83

C=8

B=1 we have:

$\displaystyle (P - Q) \cdot \left( {{P^{n - 1}} + {K_1} \cdot {P^{n - 2}} \cdot {Q^1} + {K_2} \cdot {P^{n - 3}} \cdot {Q^2} + ... + {K_{n - 2}} \cdot {P^1} \cdot {Q^{n - 2}} + {K_{n - 1}} \cdot {Q^{n - 1}}} \right) \ne {C^n} - {B^n} \Leftrightarrow $

$\displaystyle (3 - 2) \cdot ({3^{3 - 1}} + 83 \cdot {3^{3 - 2}} \cdot {2^1} + {2^{3 - 1}}) \ne {8^3} - {1^3} \Leftrightarrow $

$\displaystyle 9 + 498 + 4 \ne 511 \Leftrightarrow $

$\displaystyle 511 \ne 511$ - Jul 23rd 2013, 12:04 PMMrAwojobiRe: Proof of a Conjecture or Counter Examples
Many thanks for these counterexamples. However can I impose one last constraint i.e. (P-Q) > 1

- Jul 23rd 2013, 12:06 PMChessTalRe: Proof of a Conjecture or Counter Examples
- Jul 23rd 2013, 12:14 PMMrAwojobiRe: Proof of a Conjecture or Counter Examples
Thanks.

- Jul 23rd 2013, 12:15 PMChessTalRe: Proof of a Conjecture or Counter Examples
Well another one is:

P=4

Q=2

n=3

K1=43

C=9

B=1 - Jul 23rd 2013, 12:25 PMMrAwojobiRe: Proof of a Conjecture or Counter Examples
Thank you again. Please don't get tired of me yet but I promise this is the last constraint. P and Q should not share any common factors except the trivial unity.

- Jul 23rd 2013, 12:44 PMChessTalRe: Proof of a Conjecture or Counter Examples
- Jul 23rd 2013, 12:55 PMMrAwojobiRe: Proof of a Conjecture or Counter Examples
B>1, C and B don't share a common factor except the trivial unity are the last set of constraints. Sorry I broke my promise.