20190104, 23:38  #1 
Mar 2018
523 Posts 
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? 
20190105, 00:25  #2  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
\[2a^3+2ab^2+ac^2+a=2a^3+2b^3+c^3+1\] 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 20190105 at 00:45 

20190105, 03:35  #3 
Aug 2006
1011100100101_{2} Posts 

20190105, 03:56  #4 
Aug 2006
3×5^{2}×79 Posts 
I found that solution 'cleverly'. Brute force gives more solutions: (196, 56, 215), (265, 21, 268), (301, 26, 305), (593, 211, 669).

20190105, 17:48  #5 
Mar 2018
523_{10} Posts 
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? 
20190105, 18:31  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20190106, 05:58  #7 
Aug 2006
3·5^{2}·79 Posts 

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