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 enzocreti 2019-01-04 23:38

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?

 science_man_88 2019-01-05 00:25

[QUOTE=enzocreti;504970]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?[/QUOTE]

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.

 CRGreathouse 2019-01-05 03:35

[QUOTE=enzocreti;504970]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?[/QUOTE]

What about (a, b, c) = (26,4,27)?

 CRGreathouse 2019-01-05 03:56

I found that solution 'cleverly'. Brute force gives more solutions: (196, 56, 215), (265, 21, 268), (301, 26, 305), (593, 211, 669).

 enzocreti 2019-01-05 17:48

a slightly different 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?

 science_man_88 2019-01-05 18:31

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?[/QUOTE]
if a is even c and d need be same parity.

etc.

 CRGreathouse 2019-01-06 05:58

[QUOTE=enzocreti;505039]are there solutions with d>1?[/QUOTE]

Oh yeah. Lots. Can you find them?

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