The discussion revolves around solving a system of equations involving four variables: a, b, c, and d. The equations presented include linear and polynomial relationships, and participants are exploring potential solutions.
There is no consensus on a solution, and multiple participants express uncertainty regarding their contributions. The discussion remains unresolved.
Participants have not provided detailed workings or assumptions behind their proposed solutions, leading to ambiguity in the mathematical steps involved.
kaliprasad said:
kaliprasad said:
Thank you so much, kaliprasad for the hint and also for sharing with us of this great challenge! I could tell I've fallen in love with this problem at first sight!Hahaha...anemone said:My solution:
If we form a quartic equation as the product of two quadratic equation as follow, we have:
$\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+8x^3+23x^2+28x+12\\&\overset{I}{=}((x+1)(x+2))((x+2)(x+3))=(x^2+3x+2)(x^2+5x+6)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+2))=(x^2+4x+3)(x^2+4x+4)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+2))=(x^2+5x+6)(x^2+3x+2)\\&\overset{IV}{=}((x+2)(x+2))((x+1)(x+3))=(x^2+4x+4)(x^2+4x+3)\end{align*}$
Hence, $(a,\,b,\,c,\,d)=(3,\,5,\,2,\,6),\,(4,\,4,\,3,\,4),\,(5,\,3,\,6,\,2),\,(4,\,4,\,4,\,3)$.
But...I'm not proud of myself for this solution because...
Without the hint, I could have never solved this great challenge!