## I find this beautiful...

 Quote by morphism Yes, sorry - that was a typo. I meant t=1.
It would be interesting if one could prove that 1 is never one of the integers in the solution.

EDIT:Oh, and thank you micromass for that link. I will definitely look into it.

Blog Entries: 8
Recognitions:
Gold Member
Staff Emeritus
 Quote by Mathguy15 It would be interesting if one could prove that 1 is never one of the integers in the solution.
It will be since

$$1999=1^3+321409^3+894590^3+(-908211)^3$$

 Quote by micromass It will be since $$1999=1^3+321409^3+894590^3+(-908211)^3$$
Wow, how did you get that? Did you use some kind of computer or did you find it using some theory or theorem?

Blog Entries: 8
Recognitions:
Gold Member
Staff Emeritus
 Quote by Mathguy15 Wow, how did you get that?
Just letting a computer check the possibilities. Nothing fancy.
I didn't even program it, I found this on a list online
 Hi Mathguy15, Multiply out the terms, but first cube your 3rd term since you're going to cube it later anyway... n^3 + 30*n^2 + 300n + 1000 -n^3 + 30*n^2 - 300n + 1000 (-60*n^2) - 1 → 30*n^2 + 30*n^2 - 60*n^2 = 0 1000 + 1000 - 1 = 1999 You could do the same thing for (n+8)^3 = 1n^3 + 24*n^2 + 192n + 512, and (n-8)^3, but you'd have to replace (-60*n^2) with (-48*n^2) and 1999 with 1023. And so on... Note that you're not really dealing with cubes since they cancel out when you multiply out and collect the terms.

 Quote by Mathguy15 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. ... mathguy15(whos now 16)
Seems you're right!

http://i41.tinypic.com/107pcvr.png
 If you want to generalize further mathguy15... for any n or k in N First term: (1*k^0*n^3) + (3*k^1*n^2) + (3*k^2*n^1) + (1*k^3*n^0) = (n+k)^3 Second term: -(1*k^0*n^3) + (3*k^1*n^2) - (3*k^2*n^1) + (1*k^3*n^0) Third term: (Derived from above 2nd terms doubled) Fourth term: -1 ... then figure out the formula for those coefficients (x) above (which is almost certainly related to the Binomial Coefficients aka "the coefficients of the expansion of (x+1)^n"). Do that and then you'll have a nice little equation in 3 variables that will work for any n, k or x in N. Binomial Coefficients: 1 1 1 --> Powers of 1 1 2 1 --> Powers of 2 1 3 3 1 --> Powers of 3... 1 4 6 4 1 --> Powers of 4... ...when inserted into Polynomials. In the example you posted, then set k = 10 - AC
 Here's an example of what I was referring to in the previous post, thinking-wise. Take the form... (10 + n)^5 + (10 - n)^5 + (-100*n^4) + (-20000*n^2) + (-1) = 199999 Echoing the form used by mathguy15... v = (10 + n) w = (10 - n) x = (-100*n^4)^(1/5) y = (-20000*n'^2))^(1/5) z = -1 x = an integer for (-100*(100*s^5)^4)^(1/5); n =(100*s^5) y = an integer for (-20000*(125*s'^10)^2))^(1/5); n' = (125*s^10) Will have to see where to go from there, meaning what conclusions one might come to regarding: v^5 + w^5 + x^5 + y^5 + z^5