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 2019-01-04, 23:38 #1 enzocreti   Mar 2018 17×31 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?
2019-01-05, 00:25   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by enzocreti 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

2019-01-05, 03:35   #3
CRGreathouse

Aug 2006

3×1,987 Posts

Quote:
 Originally Posted by enzocreti 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)?

 2019-01-05, 03:56 #4 CRGreathouse     Aug 2006 174916 Posts I found that solution 'cleverly'. Brute force gives more solutions: (196, 56, 215), (265, 21, 268), (301, 26, 305), (593, 211, 669).
 2019-01-05, 17:48 #5 enzocreti   Mar 2018 17·31 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?
2019-01-05, 18:31   #6
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by enzocreti 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.

2019-01-06, 05:58   #7
CRGreathouse

Aug 2006

3·1,987 Posts

Quote:
 Originally Posted by enzocreti are there solutions with d>1?
Oh yeah. Lots. Can you find them?

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