A squares and cubes equation

[alainv...@gmail.com]

Good morning,

I propose a 'mixt equation , that is
second degree and third degree terms are used:

x,a,d integer , x being known
x^3+x^2+d^2 = (x-a)^3+a^3+(x+d)^2

For instance:
x=7
7^3+7^2+9^2 = 6^3+1^3+16^2
7^3+7^2+ ............

Best regards,

Alain

[Robert Israel]

"alainv...@gmail.com" <alainv...@gmail.com> writes:

What about it?
For x = 0 the equation is trivially true.
If x <> 0, your equation just says 3 a (x - a) = 2 d. For any integers x and
a such a and x+1 are not both odd, d = 3 a (x - a)/2 gives the solution.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[alainv...@gmail.com]

Bonjour Robert,

You'right.
I've got the soln x=a+2b ,d=3ab
I use to consider free powers expressions;
here equivalence out of squares and cubes.

Let us put p[n](u,v)= (u+v)^n -(u^n+v^n) , all integer numbers .
We may start with , say,
p[3](a,2b) = p[2](x,d) or
3*a*(2b)*((a+2b) = 2*x*d (1)
With x=(a+2b) , d=3ab
Completing with due powers we obtain:
(a+2b)^3+(a+2b)^2+(3ab)^2
=
a^3 + (2b)^3 + (a+2b+3ab)^2 .

With other values satisfying (1),
namely x=a(a+2b) , d=3b
(a+2b)^3+(a^2+2ab)^2+(3b)^2
=
a^3 + (2b)^3 + (a^2+2ab+3b)^2

I just search a simple way to build
polynomials identities,

Amicalement,
Alain