Prove please.
2006-10-09 17:20:14 · 6 answers · asked by Scott R 6 in Science & Mathematics ➔ Mathematics
No joke here.
A solution is quite possible.
2006-10-10 01:10:11 · update #1
a = 531,441
b = 6561
c = 729
d = 243
is a solution.
2006-10-10 10:50:53 · update #2
a = 109,418,989,131,512,359,209
b = 22,876,792,454,961
c = 10,460,353,203
d = 129,140,163
2006-10-10 11:00:24 · update #3
NOTE TO SHIMROD:
you might want to fix your typo...
otherwise ☺!
2006-10-10 11:51:10 · update #4
i see you fixed it just as i typed the above!
please ignore.
2006-10-10 11:52:06 · update #5
Yes, for every nonnegative integer k:
a = 3^(30k+12)
b = 3^(20k+8)
c = 3^(15k+6)
d = 3^(12k+5)
There are lots of other solutions of course.
2006-10-10 11:46:15 · answer #1 · answered by shimrod 4 · 1⤊ 0⤋
This equation definitely has solutions: consider, for instance, 2^2+3^3+1^4=2^5. Doing some brute-force searching, it looks like this has a solution whenever d is an exact power of 2 - but I don't know how to prove that. Let me think about this some more.
2006-10-10 04:32:19 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋
a^2+b^3+c^4 = k where a,b,c,k are elements of real
therefore k = d^5
therefore d =k^(1/5)
Since 5 is odd, k^(1/5) cannot be imaginary for any value of k
therefore k^(1/5) = d which is an element of real
This is not a very mathmatical proof but it works as long as you assumes that a real number to the power of an odd number will always give a real number.
2006-10-09 19:58:40 · answer #3 · answered by Morkeleb 3 · 0⤊ 3⤋
a^2 + b^3 + c^4 = d^5 : a^2 + b^3 +c^4 - d^5 = 0
2006-10-09 17:38:27 · answer #4 · answered by Freesia 5 · 0⤊ 5⤋
You're kidding me. Looks like a complicated Fermat's Last Theorem which was eventually proved by Andrew Willes. I think my math is a little bit inferior in order to even attempt to try. Nice joke though. Hahaha
2006-10-09 17:34:00 · answer #5 · answered by Hesse 3 · 2⤊ 3⤋
seems, it will have infinite solutions....
2006-10-10 11:13:14 · answer #6 · answered by m s 3 · 0⤊ 0⤋