MHB How can the variables m, n, s, and t be solved for these equations?

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion revolves around solving a set of equations involving the variables m, n, s, and t. The equations include linear, quadratic, cubic, and quartic forms, with the user expressing frustration over various unsuccessful methods attempted to find the solution. A breakthrough occurs when a participant suggests potential solutions based on the quartic equation, identifying pairs like (-2, -1, -1, 1) and (1, 2, -1, 1) as valid answers. The conversation shifts towards confirming these as the only solutions. The thread highlights the complexity of solving polynomial equations and the collaborative effort to find solutions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Hi MHB,

Problem:

Solve for reals of

$m-n-s+t=1$

$m^2+n^2-s^2-t^2=3$

$m^3-n^3-s^3+t^3=-5$

$m^4+n^4-s^4-t^4=15$

I've encountered this problem a while back and I've tried to use many methods (which include by manipulating some inequality theorems or solving them by the elimination of variables method or trying to relate the second equation and the third by multiplying the second and the third (after changing the minus sign) and let it equal to the 4th equation but all these methods have fallen apart. I am getting very tired of it and hence I hope someone could help me by giving me some hints so that I can finish the unfinished problem.

Thanks.
 
Mathematics news on Phys.org
anemone said:
Hi MHB,

Problem:

Solve for reals of

$m-n-s+t=1$

$m^2+n^2-s^2-t^2=3$

$m^3-n^3-s^3+t^3=-5$

$m^4+n^4-s^4-t^4=15$

I've encountered this problem a while back and I've tried to use many methods (which include by manipulating some inequality theorems or solving them by the elimination of variables method or trying to relate the second equation and the third by multiplying the second and the third (after changing the minus sign) and let it equal to the 4th equation but all these methods have fallen apart. I am getting very tired of it and hence I hope someone could help me by giving me some hints so that I can finish the unfinished problem.

Thanks.

I was looking at $m^4+n^4-s^4-t^4=15$ and wondered...
Could I find a solution for just this equation without going to great lengths?

Well... it might be something like $(\pm 2)^4$ combined with a couple of $(\pm 1)^4$.

And what do you know... it fits! ;)
 
Hi I like Serena, thank you very much for your reply.

You're absolutely right because based on your observation, we can tell that $(-2, -1, -1, 1)$ and $(1, 2, -1, 1)$ are two credible answers to the problem and now, I think the remaining effort is to show that these are the only two possible answers. (Smile)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top