mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > enzocreti

Reply
 
Thread Tools
Old 2019-01-04, 23:38   #1
enzocreti
 
Mar 2018

10000011112 Posts
Default An integer equation

Consider the equation

a*(2*a^2+2*b^2+c^2+1)=(2*a^3+2*b^3+c^3+1)

with a,b,c positive integers.
Are the only solutions to that equation a=1, b=1, c=1 and a=5, b=4 and c=6?
enzocreti is offline   Reply With Quote
Old 2019-01-05, 00:25   #2
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts
Default

Quote:
Originally Posted by enzocreti View Post
Consider the equation

a*(2*a^2+2*b^2+c^2+1)=(2*a^3+2*b^3+c^3+1)

with a,b,c positive integers.
Are the only solutions to that equation a=1, b=1, c=1 and a=5, b=4 and c=6?
multiplying each side through gives:

\[2a^3+2ab^2+ac^2+a=2a^3+2b^3+c^3+1\]

which then cancels down to:

\[2ab^2+ac^2+a=2b^3+c^3+1\]

which with a=b=c=1 goes to:

\[2b^3+c^3+1=2b^3+c^3+1\]

So no, there are multiple solutions. That being said, you should be able to work this out on your own, before you get taken seriously. okay sorry didn't see you listed a=c=b=1 . You could try algebraic relations between variables. then you can use them to go to univariate polynomials and apply polynomial remainder theorem.

Last fiddled with by science_man_88 on 2019-01-05 at 00:45
science_man_88 is offline   Reply With Quote
Old 2019-01-05, 03:35   #3
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

22×1,483 Posts
Default

Quote:
Originally Posted by enzocreti View Post
Consider the equation

a*(2*a^2+2*b^2+c^2+1)=(2*a^3+2*b^3+c^3+1)

with a,b,c positive integers.
Are the only solutions to that equation a=1, b=1, c=1 and a=5, b=4 and c=6?
What about (a, b, c) = (26,4,27)?
CRGreathouse is offline   Reply With Quote
Old 2019-01-05, 03:56   #4
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

22×1,483 Posts
Default

I found that solution 'cleverly'. Brute force gives more solutions: (196, 56, 215), (265, 21, 268), (301, 26, 305), (593, 211, 669).
CRGreathouse is offline   Reply With Quote
Old 2019-01-05, 17:48   #5
enzocreti
 
Mar 2018

17·31 Posts
Default a slightly different equation

what instead about the equation:


a(2*a^2+2*b^2+c^2+d^2)=(2*a^3+2*b^3+c^3+d^3)?


Again a=1,b=1,c=1,d=1 is a solution
a=5, b=4, c=6, d=1 is another solution


are there solutions with d>1?
enzocreti is offline   Reply With Quote
Old 2019-01-05, 18:31   #6
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

836910 Posts
Default

Quote:
Originally Posted by enzocreti View Post
what instead about the equation:


a(2*a^2+2*b^2+c^2+d^2)=(2*a^3+2*b^3+c^3+d^3)?


Again a=1,b=1,c=1,d=1 is a solution
a=5, b=4, c=6, d=1 is another solution


are there solutions with d>1?
if a is even c and d need be same parity.
if a=b then ac^2+ad^2=c^3+d^3

etc.
science_man_88 is offline   Reply With Quote
Old 2019-01-06, 05:58   #7
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

22×1,483 Posts
Default

Quote:
Originally Posted by enzocreti View Post
are there solutions with d>1?
Oh yeah. Lots. Can you find them?
CRGreathouse is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
k*b^n+/-c where b is an integer greater than 2 and c is an integer from 1 to b-1 jasong Miscellaneous Math 5 2016-04-24 03:40
Possible solutions to an equation: Vijay Math 6 2005-04-14 05:19
Time to prp equation jasong Math 2 2005-03-11 03:30
Request for an equation. Logic Programming 1 2004-09-19 21:26
Recurrence Equation jinydu Puzzles 6 2004-05-15 14:02

All times are UTC. The time now is 11:42.

Thu Oct 29 11:42:44 UTC 2020 up 49 days, 8:53, 1 user, load averages: 1.73, 1.76, 1.86

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.