# Proof of a Conjecture or Counter Examples

• Jul 23rd 2013, 07:56 AM
MrAwojobi
Proof 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, 09:00 AM
MrAwojobi
Re: Proof of a Conjecture or Counter Examples
Can anyone please prove or disprove this?
• Jul 23rd 2013, 10:53 AM
ChessTal
Re: Proof of a Conjecture or Counter Examples
Quote:

Originally Posted by MrAwojobi
{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.

For:
P=2
Q=1
n=3
K1=4
K2=7
C=3
B=2 we have:

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

$\displaystyle (2 - 1) \cdot ({2^{3 - 1}} + 4 \cdot {2^{3 - 2}} \cdot {1^1} + 7 \cdot {1^{3 - 1}}) \ne {3^3} - {2^3} \Leftrightarrow$

$\displaystyle 4 + 8 + 7 \ne 19 \Leftrightarrow$

$\displaystyle 19 \ne 19$

Obviously the conjecture can't be true.