OK, if the solution is to be limited to integers, that's easy. 97 - x must be > 0 or we'd get into complex solutions. X must be an a^4, so the 4th root yields an integer. a=4 would already be to large but a=3 works: 3^4 = 81 and 97-81 = 16 = 2^4. First root: 3, second root 2 Sum 5. a= 2 also works reversing the value of the two roots still yielding a sum of 5. => 2 integer solutions: x=16 and x=81 qed.
I don’t see how you concluded a-2 and a-3 were factors except by intuiting that 97 is the sum of two numbers which are 2 perfect quartic roots. And since you can restate the problem as: find all x,y such that x,y are natural numbers, x+y=97, and x^.25 + y^.25 = 5, then brute force is faster than the binomial expansion, 2 polynomial divisions, and one quadratic solve. Even when replacing 97 with 195857 and 5 with 34, brute force will be faster.
OK, if the solution is to be limited to integers, that's easy. 97 - x must be > 0 or we'd get into complex solutions. X must be an a^4, so the 4th root yields an integer. a=4 would already be to large but a=3 works: 3^4 = 81 and 97-81 = 16 = 2^4. First root: 3, second root 2 Sum 5. a= 2 also works reversing the value of the two roots still yielding a sum of 5.
=> 2 integer solutions: x=16 and x=81 qed.
Fantastic bro 👏
I don’t see how you concluded a-2 and a-3 were factors except by intuiting that 97 is the sum of two numbers which are 2 perfect quartic roots.
And since you can restate the problem as: find all x,y such that x,y are natural numbers, x+y=97, and x^.25 + y^.25 = 5, then brute force is faster than the binomial expansion, 2 polynomial divisions, and one quadratic solve.
Even when replacing 97 with 195857 and 5 with 34, brute force will be faster.
Tricky is overrated.