All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Nonnegative Integer Decision Poser (Posted on 2016-03-22)
Consider this system of equations:
A+B*C = (A+B)2, and:
B+C*A = (B+C)2, and:
C+A*B = (C+A)2, and:

It is observed that A=B=C=0 trivially satisfies the above system of equations.

Does there exist any further nonnegative integer solutions to the above system of equations?

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 1 of 1
First, look for any other solutions with 0.

If A=0 then the system becomes:
B*C = B^2
B = (B+C)^2
C = C^2

C is either 0 or 1.  If C=0 then B=0, yielding the trivial solution (0,0,0).  If C=1 then B = B^2 = (B+1)^2, which has no solutions.

Then the only solution (A,B,C) with at least one zero is (0,0,0).

If A=B then the original system reduces to:
A*(1+C) = (2A)^2
A*(1+C) = (A+C)^2
C+A^2 = (A+C)^2

The first two imply 2A = A+C, or A=C.  Then A=B=C, which implies they all equal 1/3 or 0, but 1/3 is not integral.

To try to find remaining nonnegative integer solutions assume without loss of generality 1<=A<B<C.

The second equation states B + C*A = (B+C)^2 = B^2 + 2BC + C^2.  But from the inequality B<B^2 and A*C<C^2.  Therefore B + C*A < B^2 + 2BC + C^2 and the equality has no solutions.

The only nonnegative integer solution is (0,0,0).

 Posted by Brian Smith on 2016-03-22 13:54:30

 Search: Search body:
Forums (0)