// // For my colleagues in Shell with a lot of love,  (and  with a lot of time now since no commuting, cause COVID) // Code is commented to explain how to solve the meme  (https://preview.redd.it/p92108lekoq11.jpg?width=367&format=pjpg&auto=webp&s=e0c84917c3d7e130cad06f9ab5a85634b0c88cfb) // // x/(y+z) + y/(x+z) + z/(x+y) = 4 // // This is the smallest solution: // x=4373612677928697257861252602371390152816537558161613618621437993378423467772036 // y=36875131794129999827197811565225474825492979968971970996283137471637224634055579 // z=154476802108746166441951315019919837485664325669565431700026634898253202035277999 // // Paste in the site below to execute this code see this result, also read the comments here to understand.  // The last part of the prints() after executed shows you the solution above. // http://magma.maths.usyd.edu.au/calc/ // Eduardo Ruiz Duarte  // [email protected] // // First we define our environment for our "problem" R<x,y,z> := RationalFunctionField(Rationals(),3); problem := ((x/(y+z) + y/(x+z) + z/(x+y)) - 4) ; // first note that we know a point after some computation (-1,4,11) that // works but has a negative coordinate, the following function returns 0, which means that  // (x/(y+z) + y/(x+z) + z/(x+y)) - 4 = 0    (just put the -4 in the other side) Evaluate(problem,[-1,4,11]); // after the previous returned 0 , we know the point fits, we continue. // we multiply by all the denominators of "problem" to get a polynomials problem*Denominator(problem); // we obtain a polynomial without denominators x^3 - 3*x^2*y - 3*x^2*z - 3*x*y^2 - 5*x*y*z - 3*x*z^2 + y^3 - 3*y^2*z - 3*y*z^2 + z^3 // We see is cubic, three variables, and every  term has the same degree (3) , therefore this is a cubic  // homogeneous curve,  we know there is a point which is not the solution we want // the point (-1,4,11) fits in the original "problem" so it should fit in this new curve without denominators too (since no denominator becomes 0) // We transform this equation to a "curve" in Projecive space of dimension 2 P2<x,y,z> := ProjectiveSpace(Rationals(),2); C := Curve(P2,x^3 - 3*x^2*y - 3*x^2*z - 3*x*y^2 - 5*x*y*z - 3*x*z^2 + y^3 - 3*y^2*z - 3*y*z^2 + z^3); // fit the point to the curve C (no error is returned) Pt := C![-1,4,11]; // Since all cubic homogeneous curve with at least one point define an elliptc curve, we can transform  // this curve C to an elliptc curve form and just like in cryptography, we will add this known point (mapped to the corresponded curve) // with itself until we get only positive coordinates and go back to C (original Problem) // Below, E is the curve, f is the map that maps   Points f:C -> E  (C is our original curve without denominators, both curves C,E are equivalent  // but in E we can "Add points" to get another point of E. // and with f^-1 we can return to the point of C which is our original solution E,f := EllipticCurve(C); //g is the inverse g:E->C  , f:C->E     so g(f([-1,4,11]))=[-1,4,11] g := f^-1; // We try adding the known point Pt=[-1,4,11] mapped to E, 2..100 times // to see if when mapped back the added point to C gives positive coordinates //, this is 2*Pt, 3*Pt, ...., 100*Pt  and then mapping back to C all these. for n:= 1 to 100 do // we calculate n times the point of C, known [-1,4,11] but mapped (via f) inside E (where we can do the "n times")      nPt_inE:=n*f(Pt); // we take this point on E back to C via f^-1  (which we renamed as g)     nPt_inC:=g(nPt_inE); //We obtain each coordinate of this point to see if is our positive solution, // here MAGMA scales automatically the point such as Z is one always 1,  // so it puts the same denominators in X,Y, so numerators of X,Y are our  //solutions and denominator our Z,  think of  P=(a/c,b/c,1)   then c*P=(a,b,c)     X := Numerator(nPt_inC[1]);     Y := Numerator(nPt_inC[2]);     Z := Denominator(nPt_inC[1]); printf "X=%o\nY=%o\nZ=%o\n",X,Y,Z; // We check the condition for our original problem.   if ((X gt 0) and (Y gt 0)) then        printf("GOT IT!!! x=apple, y=banana, z=pineapple, check the above solution\n");      break;   else      printf "Nee, some coordinate was negative above, I keep in the loop\n\n";   end if; end for;    // We check the solution fits in the original problem if Evaluate(problem, [X,Y,Z]) eq 0 then     printf "I evaluated the point to the original problem and yes, it worked!\n"; else     printf "Mmm this cannot happen!\n"; end if;

###Test//R<x,y,z> := RationalFunctionField(Rationals(),3); //problem := ((x/(y+z) + y/(x+z) + z/(x+y)) - 6) ; //Evaluate(problem,[-23, -7, 3]); P2<x,y,z> := ProjectiveSpace(Rationals(),2); C := Curve(P2,x^3 - 5*x^2*y - 5*x^2*z - 5*x*y^2 - 9*x*y*z - 5*x*z^2 + y^3 - 5*y^2*z - 5*y*z^2 + z^3); Pt := C![-23, -7, 3]; E,f := EllipticCurve(C); g := f^-1; for n:= 1 to 100 do      nPt_inE:=n*f(Pt);     nPt_inC:=g(nPt_inE);     X := Numerator(nPt_inC[1]);     Y := Numerator(nPt_inC[2]);     Z := Denominator(nPt_inC[1]);   if ((X gt 0) and (Y gt 0)) then        printf("GOT IT!!! x=apple, y=banana, z=pineapple, check the above solution\n");       printf "n=%o\nX=%o\nY=%o\nZ=%o\n",n,X,Y,Z;   end if; end for;###Test//if Evaluate(problem, [X,Y,Z]) eq 0 then     //printf "I evaluated the point to the original problem and yes, it worked!\n"; //else     //printf "Mmm this cannot happen!\n"; //end if;