My Math Forum
>
College Math Forum
>
Number Theory
(ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d)
User Name
Remember Me?
Password
Register
Home
Forums
Number Theory
Number Theory Math Forum
6
Thanks
Top
All
This Page
Page 2 of 2
<
1
2
LinkBack
Thread Tools
Display Modes
January 10th, 2017, 04:43 PM
#
11
Denis
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada
Posts: 10,654
Thanks: 697
Quote:
Originally Posted by
Naveenchandra Kumar
It is (5,7,
, .... (117, 228, 23
.
Last edited by Naveenchandra Kumar; January 10th, 2017 at 08:34 PM. Reason: Somehow smiley is sitting over 8! always.
If last digit in brackets is 8, leave a space: (238 )
Thanks from
Naveenchandra Kumar
Page 2 of 2
<
1
2
My Math Forum
>
College Math Forum
>
Number Theory
Tags
ax3
,
ay3
,
az3
,
bx2
,
by2
,
bz2
,
integer
,
solution
«
multiple of a number

Factorization . Is this possible ?
»
Thread Tools
Show Printable Version
Email this Page
Display Modes
Linear Mode
Switch to Hybrid Mode
Switch to Threaded Mode
Contact

Home

Forums

Cryptocurrency Forum

Top
Copyright © 2017 My Math Forum. All rights reserved.
LinkBack
LinkBack URL
About LinkBacks