Hi all,Three integers a, b and c such that andFind all the possible values of .weatherwax

a,b,c > 0:First equation => a = 1 and b or c is equal to 0.{c = 0 => b = sqrt(2012) which is not integer} => only pos sol is a = 1, b = 0, c = 2012/2.Observations for negative integers:1. b is positive implies c is negative.2. a is odd and b is even.

hint: 2011 is a prime number

My first guess is N^2 - 2013, but it might be a subset of that.

1006^2

QuoteOriginally posted by: outruna=1, b=0, c=1006 is a solution, or are all integers supposed to be >0 ?The original question says integers, so I assumed a,b,c can be positive and negative. But the value in question is independent of sign, as there is a symmetry a->-a, b->-b, c->-c.

QuoteOriginally posted by: outruna=1, b=0, c=1006 is a solution, or are all integers supposed to be >0 ?The above solution or with the opposite signs.Following the wileysw's hint: (b^2 + 2ac) - (a^2 + 2bc) = (b - a)*(b + a - 2c) = 2011. Since 2011 is a prime number, there are four possible cases:1) b - a = 1, b + a - 2c = 20112) b - a = 2011, b + a - 2c = 13) b - a = 1, b + a - 2c = -20114) b - a = 2011, b + a - 2c = -1I calculated the square of the second equality (hence it's enough to consider only case 1 and 2) and did some substitutions and also got only one integer solution, 1006^2.

Now do it for a,b,c Reals (and not Necessarily integers).

AVt, i thought it would be messy algebra for your question, but it was a pleasant surprise to find the following general conclusion:a^2+2*b*c=x, b^2+2*c*a=y, c^2+2*a*b=z have real solutions iff x, y, z are real and x+y+z>=0.to prove the "if" direction (the other direction is trivial), notice one can "diagonalize" ("linearize" might be more appropriate) by using the cube root of 1: omega=exp(2*i*pi/3), which satisfies 1+omega+omega^2=0, so one can easily get the equivalent equations:(a+b+c)^2=(x+y+z), (a+omega*b+omega^2*c)^2=(x+omega^2*y+omega*z), (a+omega^2*b+omega*c)=(x+omega*y+omega^2*z);now that they are reduced to a linear system, one can explicitly write down the solution, but the hint/spolier to finish the last step of the proof is to note omega and omega^2 are complex conjugates.(i was not able to find a general theorem, but suspect it should be related to representation theory of quadratic form)back to the original question, one simple gets a range: c^2+2*a*b >=-2013

Yes, but I did the Algebra (can you diagonalize simultanously here?), having Mapleat hand and for the range it is more easy as for exact solutions:Write c = (1-a^2)/b/2. Then we are left with b^3+a-a^3 = 2012*b (case b=0 is easy).Then the cubic in 'a' has solutions, depending on b (actually using the discriminantone can see, there is always exactly 1 real solution + 2 complex except for 3 pointsb, which give 1 double root each).Feed that into c^2+2*a*b, differentiate w.r.t. b, set it to 0 and solve it for b.That has only 1 real solution, b0 = -sqrt((6036+2*sqrt(12150471)))/3 ~ -38.01and feeding again one finally arrives at -2013 as local extremum.Which is a minimum (for example take b=0 or b=1 or any other simple input).For that one has a = -3*b0*(sqrt(12150471)/6042-2011/6042) ~ 27.84Of course all that would be a mess doing the lengthy expressions by hand, Maple usesan abbreviation 'RootOf' instead, which may be similar to your approach of unit roots.

AVt, that does not seem fun...the solutions have much nicer forms, e.g., the real solutions: (note the complex conjugated terms)a = {+/- sqrt(x+y+z) +/- [omega*sqrt(x+omega*y+omega^2*z) + omega^2*sqrt(x+omega^2*y+omega*z)]}/3,where the sign convention of sqrt (a long discussion) is to take the root with the positive real part. b and c follow from the cyclic symmetry (though be careful with the signs)