Why qxy(x + y)[(x + y)^2 - xy)]^p = (x + y)^q – x^q – y^q cannot exist if q = 2p + 3 and when p > 2?

Question

Here x and y are integers greater than zero and p is an integer greater than or equal to zero.

The equation is valid when p = 0, 1 and 2 but it fails when p > 2. Why?

Please help. Thanks!

x, y must be integers > 0
p must be an integer ≥ 0

There's someone here known as Cool Guy with only 4 answers and made as of July 25, 2007 but calls everybody an IDIOT! What a pitiful creature!!

Answer 1

I'm not very good in math but let me help a bit.

When we expand the binomial (x + y)^q and subtract x^q and y^q from it, we can always factor out q from the difference if q is a prime number!

Well, you can observe that this is not possible when q is a composite number. Am I right?

That's why the equation cannot exist for composite q's.

Yeah, I opened Cool Guy's Q&A and I found nothing but bullshit!

Answer 2

Consider (x+y)^q - x^q - y^q. Since q is odd we know (x+y) is a factor of x^q + y^q and hence of (x+y)^q - x^q - y^q.
x^q + y^q = (x+y) [x^(q-1) - x^(q-2)y + ... + y^(q-1)]
So (x+y)^q - x^q - y^q = (x+y) [(x+y)^(q-1) - x^(q-1) + x^(q-2)y - ... - y^(q-1)]
= (x+y) [x^(q-1) + (q-1)x^(q-2)y + ... + (q-1)Cn x^(q-n-1) y^n + ... + y^(q-1) - x^(q-1) + x^(q-2)y - ... - y^(q-1)]
= (x+y) [q x^(q-2) y + ... + ((q-1)Cn - (-1)^n) x^(q-n-1) y^n + ... + (q+1) x y^(q-2)]
= xy (x+y) [q x^(q-3) + ... + ((q-1)Cn - (-1)^n) x^(q-n-2) y^(n-1) + ... + q y^(q-3)]
where n goes from 1 to q-2 (with the first and last terms explicitly written out above).

So the answer is that q[(x+y)^2 - xy]^p is contrived to equal [q x^(q-3) + ... + ((q-1)Cn - (-1)^n) x^(q-n-2) y^(n-1) + ... + q y^(q-3)] for q = 3, 5, 7 but fails for larger values of q. It will always match the first two and last two terms, but fails in the middle.

e.g. for p = 3, q = 9:
short poly = 9[(x+y)^2 - xy]^3 = 9(x^2 + xy + y^2)^3
= 9x^6 + 27x^5y + 54x^4y^2 + 63x^3y^3 + 54x^2y^4 + 27xy^5 + 9y^6
long poly = 9x^6 + (8C2 - 1)x^5y + (8C3+1)x^4y^2 + (8C4-1)x^3y^3 + (8C5+1)x^2y^4 + (8C6-1)xy^5 + 9y^6
= 9x^6 + 27x^5y + 57x^4y^2 + 69x^3y^3 + 57x^2y^4 + 27xy^5 + 9y^6

Answer 3

IN LAMENS TERMS x^q<8cos+xy\yx=p

p=7.5

IDIOT, LEARN TO COUNT.