For anyone fascinated by powers and integers, there’s no shortage of problems to tackle, whether by ingenious logic or massive computer search.

In 1769, while thinking about the problem now known as Fermat’s last theorem, Leonhard Euler (1707–1783) proposed an intriguing variant.

A century earlier, Pierre de Fermat (1601–1665) had proved that no set of positive integers, *a*, *b*, and *c*, satisfies the equation *a*^{4} + *b*^{4} = *c*^{4} in the same way that numbers such as 3, 4, and 5 satisfy the more familiar equation *x*^{2} + *y*^{2} = *z*^{2}. Euler conjectured that the equation *a*^{4} + *b*^{4} + *c*^{4} = *d*^{4} would also have no integer solutions.

Indeed, Euler went further. He proposed that for every integer greater than 2, the sum of *n* – 1 *n*th powers of positive integers cannot itself be an *n*th power.

In 1966, L.J. Lander and T.R. Parkin disproved Euler’s generalized conjecture when they found a counterexample for the case *n* = 5. The equation *a*^{5} + *b*^{5} + *c*^{5} + *d*^{5} = *e*^{5} is true when *a* = 27, *b* = 84, *c* = 110, *d* = 133, *e* = 144.

In 1986, Noam D. Elkies of Harvard University found the first counterexample that proved Euler’s conjecture was wrong for the case *n* = 4. The equation is true when *a* = 2,682,440, *b* = 15,365,639, *c* = 18,796,760, and *d* = 20,615,673.

Elkies discovered the counterexample by a method that combined theoretical reasoning and a relatively short computer search. He converted the problem into an equivalent mathematical form that enabled him to pick out candidates likely to satisfy Euler’s equation. A few mathematicians had tried a similar approach previously, but either they had given up too soon or they had been thinking in terms of proving rather than disproving the conjecture.

Once Elkies found the first counterexample, he was able to prove that there are infinitely many, each consisting of enormously large numbers. What Elkies didn’t know at the time was whether his initial solution was the smallest one.

When Roger Frye, then at Thinking Machines in Cambridge, Mass., heard news of Elkies’s achievement, he tackled the problem himself. Using hints supplied by Elkies to shorten the search, he wrote a computer program to look for smaller solutions.

Working at night in his spare time, Frye used Connection Machine computers to find the smallest positive integers that fit the equation. His exhaustive computer search showed that *a* = 95,800, *b* = 217,519, *c* = 414,560, and *d* = 422,481.

Subsequent computer searches turned up additional examples. In 2000, for *d* less than 2.1 x 10^{7}, the total stood at seven, with values for *d* of 422,481, 2,813,001, 8,707,481, 12,197,457, 16,003,017, 16,430,513, and 20,615,673.

That still leaves lots of problems and conjectures about sums of powers yet unsolved. For example, no one has found a counterexample to Euler’s conjecture for any power greater than 5. Can a sixth power be expressed as the sum of five sixth powers?

Anyone with a computer interested in investigating “minimal equal sums of like powers” can join in the fun (see http://euler.free.fr/index.htm). Recently, such a computer search turned up a second solution for the case *n* = 5: 85359^{5} = 85282^{5} + 28969^{5} + 3183^{5} + 55^{5}.

And what about the sum of *four* fourth powers expressed as a fourth power? The list goes on and on.

**Puzzle of the Week**

A home has three clocks. On Jan. 1 at midnight, all three show the correct time. But only the first clock keeps perfect time. The second clock loses a minute every day, and the third one gains a minute every day. If the clocks were to continue in this way, how long would it be before all of them show the correct time again at the same time?

For the answer, go to http://www.sciencenewsforkids.org/articles/20031022/PuzzleZone.asp.