- #1
Mathguy15
- 68
- 0
I was studying (yet another) number theory problem, described here:
Prove that the equation x^3+y^3+z^3+t^3=1999 has infinitely many solutions (x,y,z,t) in integers.
I found a way of constructing these solutions, which I will describe right now: Consider the quadruple of real numbers of the form (10+n,10-n,-(60n^2)^1/3,-1), where n is an integer. The sum of the cubes of these real numbers, for arbitrary n, is always 1999. Let n be of the form s^3*60, where s is any integer. Then the quadruple described above becomes an integer quadruple. Since there are infinitely many such n, there are infinitely many integer solutions to the equation.
I found this fascinating! Can you all find an alternate way of solving the problem? I'd be interested.
Thanks
mathguy15(whos now 16)
Prove that the equation x^3+y^3+z^3+t^3=1999 has infinitely many solutions (x,y,z,t) in integers.
I found a way of constructing these solutions, which I will describe right now: Consider the quadruple of real numbers of the form (10+n,10-n,-(60n^2)^1/3,-1), where n is an integer. The sum of the cubes of these real numbers, for arbitrary n, is always 1999. Let n be of the form s^3*60, where s is any integer. Then the quadruple described above becomes an integer quadruple. Since there are infinitely many such n, there are infinitely many integer solutions to the equation.
I found this fascinating! Can you all find an alternate way of solving the problem? I'd be interested.
Thanks
mathguy15(whos now 16)