Mark Dettinger's Conjecture Mark Dettinger's Conjecturea^n + b^n = c^n has no integer solutions for n>2.In other words, I think, Fermat's Last Theorem is only the first and simplest one of an infinite cascade of theorems.
a^n + b^n + c^n = d^n has no integer solutions for n>3.
(a^n + b^n + c^n + d^n = e^n has no integer solutions for n>4.)
a^n + b^n + c^n + d^n + e^n = f^n has no integer solutions for n>5.
a^n + b^n + c^n + d^n + e^n + f^n = g^n has no integer solutions for n>6.
...
What do you think? Do you know a second counter-example?
By the way, for a^4 + b^4 + c^4 = d^4 there's no one with d<=220000.
For a^5 + b^5 + c^5 + d^5 = e^5 there's no second one with e<=765.
Noam Elkies found:
2682440^4 + 15365639^4 + 18796760^4 = 20615673^4
New York Times carried his solution.
Then Roger Fry (Thinking Machines Corporation) found by brute force search
95800^4 + 217519^4 + 414560^4 = 422481^4 and this is the smallest integer
solution.
Noam published a paper showing how his and Roger's solutions may be obtained
from certain conductors or pencils of elliptic curve and surface intersections
and rational points on these surfaces. Actually I think Noam started with the
Fauquamberque parametrization and worked from there.
How many terms do you have to add to make a n-th power?
For n=2 you need 2 terms (3^2 + 4^2 = 5^2).
For n=3 you need 3 terms (3^3 + 4^3 + 5^3 = 6^3).
For n=4 you need 3 terms (95800^4 + 217519^4 + 414560^4 = 422481^4).
How many terms do you need for n=5? 3 or 4?
Let f(n) denote the number of terms you need for a n-th power.
Now my new conjecture is:
f(n) is monotonically increasing.
(Note, that the original conjecture was f(n) >= n.)
Another generalization of Fermat's Last Theorem.