How find the value of the $x+y$

Question:

let $x,y\in \Bbb R $, and such $$\begin{cases} 3x^3+4y^3=7\\ 4x^4+3y^4=16 \end{cases}$$

Find the $x+y$

This problem is from china some BBS

My idea: since $$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$ $$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$

then I can't Continue

Solution 1:

Assuming the question is typed correctly as shown, there are two unique real solutions. Let $$\begin{align*} u(z) &= -983749-111132z^3+786432z^4+71442z^6-196608z^8-20412z^9+18571z^{12}, \\ v(z) &= -178112-351232z^3+186624z^4+301056z^6-34992z^8-114688z^9+18571z^{12}. \end{align*}$$ These polynomials have exactly two distinct real roots; let $r(u,+)$, $r(u,-)$ be the positive and negative real roots of $u$, and $r(v,+)$, $r(v,-)$ be the positive and negative real roots of $v$, respectively. Then $$(x,y) \in \{(r(u,-),r(v,+)), (r(u,+),r(v,-))\}$$ are the desired solutions. The sum $x+y$ can then be expressed by the solution to a third polynomial $$f(z) = 819447-537600z-8998752z^3+3291428z^3+22132992z^4-17875200z^5+3163146z^6+1042512z^8-437500z^9+18571z^{12},$$ for which there are again two real roots, both positive. All of these polynomials are irreducible. So I highly doubt that this is a problem that can be reasonably solved by hand.

Solution 2:

Here comes a drawing from a credible (Delphi Pascal) source (-:

Viewport:

  xmin := -4; xmax := +4;
  ymin := -4; ymax := +4;

EDIT. A parameter representation for the curve $\;3x^3+4y^3=7\;$ is: $$ x(t) = -\sqrt[3]{\frac{t}{3}} \qquad ; \qquad y(t) = \sqrt[3]{\frac{t+7}{4}} $$ This turns the problem into seeking zeroes of a one-dimensional function: $$ f(t) = 4 x^4(t) + 3 y^4(t) - 16 $$ Standard numerical methods can be employed for this purpose. Newton-Raphson and Regula Falsi have been successful here. Our (double precision) results are: $$ t = 3.56874527617414 \quad \Longrightarrow \quad (x,y) = ( -1.05957433921527 , +1.38246606572819 ) \\ t = -8.23503156460535 \quad \Longrightarrow \quad (x,y) = ( +1.40017183631184 , -0.675884813969605 ) $$ Giving, respectively: $$ x+y = 0.322891726512912 \\ x+y = 0.724287022342236 $$ ALGORITHM (Delphi Pascal) :

program RF;

type
  funktie = function(t : double) : double;

function regula_falsi(d1,d0,eps : double; F : funktie) : double;
{
  Regula Falsi
}
var
  OK : boolean;
  d2 : double;
begin
  d2 := d0;
  while abs(F(d2)) > eps do
  begin
    d2 := d1 - F(d1)*(d0 - d1)/(F(d0) - F(d1));
    if F(d2) = 0 then Break;
    OK := (F(d2)F(d1) < 0);
    if not OK then
    begin
      d1 := d2;
    end else begin
      d0 := d1; d1 := d2;
    end;
    Write(d2,' : ',F(d2),' ; Press Enter'); Readln;
  end;
  regula_falsi := d2;
end;

function power(x,r : double) : double;
var
  M : double;
begin
  M := 0;
  if x > 0 then
    M := exp(rln(abs(x)));
  if x < 0 then
    M := -exp(r*ln(abs(x)));
  power := M;
end;

function original(t : double) : double;
{
  Function itself
}
var
  x,y,f : double;
begin
  x := -power(t/3,1/3);
  y :=  power((t+7)/4,1/3);
  f := 4*sqr(sqr(x))+3*sqr(sqr(y))-16;
  original := f
end;

procedure Calculate(x1,x2 : double);
const
  eps : double = 1.E-14;
var
  t,x,y,t1,t2 : double;
begin
  t1 := 3*sqr(x1)*x1;
  t2 := 3*sqr(x2)*x2;
  t := Regula_Falsi(t1,t2,eps,original);
  Writeln(t);
  x := -power(t/3,1/3);
  y :=  power(t/4+7/4,1/3);
  Writeln(x,' +',y,' =',x+y); 
end;

begin
  Calculate( 1.0, 1.5);
  Writeln;
  Calculate(-1.5,-0.5);
end.

Solution 3:

Defining $z:= x+y$, the equations become $$\begin{cases} 3x^3+4(z-x)^3=\phantom{1}7\\ 4x^4+3(z-x)^4=16 \end{cases}$$

Using the method of resultants (which could conceivably be done "by hand", but I used Mathematica because life's too short) to eliminate $x$, we confirm @heropup's result:

$$\begin{align} 0 &= 819447 - 537600 z - 8998752 z^2 + 3291428 z^3 + 22132992 z^4 \\ &- 17875200 z^5 + 3163146 z^6 + 1042512 z^8 - 437500 z^9 + 18571 z^{12} \quad (\star) \end{align}$$

Solving an irreducible $12$th-degree polynomial equation is tricky at best. Again using Mathematica, the two real roots of $(\star)$ are found to be $$z=0.322892\dots \quad\text{and}\quad z=0.724287\dots$$

If we abstract the equations like this ... $$\begin{cases} a_1 x^3 + b_1 (z-x)^3=c_1\\ a_2 x^4 + b_2 (z-x)^4=c_2 \end{cases}$$ ... then the corresponding resultant is ... $$\begin{align}0 &= c_1^4\;(a_2 + b_2)^3 - c_2^3\;(a_1 - b_1)^4 \\ &- 12 z \; (a_1 - b_1)^2 \; (a_1 b_2 + a_2 b_1)\; c_1 c_2^2 \\ &+ \cdots \\ &- 24 z^5\;(a_1 b_2 + a_2 b_1) \; (a_1 a_2 b_1 + 2 a_2 b_1^2 + 2 a_1^2 b_2 + a_1 b_1 b_2) \; c_1 c_2 \\ &+ \cdots \\ &- 4 z^9 \; ( a_1 b_2 + a_2 b_1)^3 \;c_1 \\ &+ z^{12}\; ( a_1^4 b_2^3 + a_2^3 b_1^4 ) \end{align}$$

Observations:

• A number of coefficients vanish under the assumption $a_1 b_2 + a_2 b_1 = 0$, although this doesn't really help much, because the polynomial remains irreducible still and degree $12$.

• We can reduce the degree to $9$ with the relation $a_1^4 b_2^3 + a_2^3 b_1^4 = 0$, but that's also not particularly helpful.

• We can eliminate the constant term if $c_1 = |a_1 - b_1| c^3$ and $c_2 = (a_2 + b_2) c^4$ for some $c$. In the original problem, with $a_1 = b_2 = 3$ and $b_1 = a_2 = 4$, this gives $c_1 = c^3$ and $c_2 = 7 c^4$. If $b_2$ had a sign-change to $-3$, then $c_1 = c^3$ and $c_2 = c^4$, which for $c=2$, give $8$ and $16$ ... almost the $7$ and $16$ of the original problem. Even so, polynomial $(\star)$ only reduces to $11$th degree.

If there's something remarkable about the solutions to this problem, I'm not seeing it.