 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Cubic coefficients (Posted on 2020-11-02) The coefficients a,b,c of a polynomial f:R->R, f(x)=x3+ax2+bx+c are mutually distinct integers and different from zero. Furthermore, f(a)=a3 and f(b)=b3. Determine a,b and c.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: Possible solution; rest of a solution | Comment 2 of 3 | (In reply to Possible solution; not a proof by Larry)

The equation a^3 + ab - ab^2 - b^2 = 0 can be factored.

Group in to (a^3-ab^2) + (ab-b^2) = 0.  Then factor each group:
a*(a-b)*(a+b) + b*(a-b) = 0
(a-b)*[a*(a+b)+b] = 0

Then either a-b = 0 or a*(a+b)+b = 0.  The first equation has solutions not permitted by the problem's conditions, therefore a*(a+b)+b = 0.

Now solve for b:
a*(a+b)+b = 0
a^2 + (a+1)*b = 0
b = -a^2/(a+1)
b = (1-a^2)/(a+1) - 1/(a+1)
b = 1 - a - 1/(a+1)

To be integers, a+1 must be a factor of -1.  Therefore a+1=1 or a+1=-1.  Then a=0 or -2.  a=0 can be discarded which leaves a=-2.

Then b = 1 - (-2) - 1/(-2+1) = 1 + 2 + 1/1 = 4
Then (-2)^3 + (-2)*4 + c = 0, which makes c = -(-8 + -8) = 16.

Then the solution to the problem is (a,b,c) = (-2, 4, 16).

 Posted by Brian Smith on 2020-11-02 22:41:07 Please log in:
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