**DIOPHANTINE FUN**

Here is an interesting conjecture that I posed some time ago in the sci.math newsgroups.

For background, let us look at a class of equations called Diophantine equations. This is the set of equations whose solutions are sought in terms of integers. Actually, Diophantus, an ancient Greek mathematician and the originator of these equations, looked at a more general set of equations in which the solutions were rational numbers (i. e., numbers of the form k/j where k and j are integers) but we will limit our conjecture to integers. Probably the most famous diophantine equation is known as Fermat’s Last Theorem (or the FLT for short). Fermat was a 17th century mathematician/physicist who dabbled in numbers. Basically, Fermat’s Last Theorem says that when n > 2, there are no integers k, l, and m that satisfy the equation

k^{n} + l^{n} = m^{n}.

Fermat noted that he had a very simple solution to this but the margin of the book was not sufficient to contain it. Mathematicians have searched for the solution to this problem for centuries with little success. But, FLT made the news a few years ago when Andrew Wiles, a Princeton mathematician, claimed a proof. His proof came under close scrutiny by the Mathematics community. Some people still say that there were flaws in Wiles' work. However, the proof has been generally accepted by the Mathematics community.

Of course, many of us are familiar with this equation when n = 2 because Pythagorus’ famous equation is a special case of this. As a refresher, Pythagorus was also an ancient Greek mathematician and his equation says that the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse. Thus integers such as 3, 4, and 5 are said to form a Pythagorean triplet since 3^{2} + 4^{2} = 5^{2}. Thus, if you have a right triangle with two sides equal to 3 and 4 units respectively, the third side will be of length 5 units.

There are many sets of numbers that satisfy this equation. But, this set of numbers, 3, 4 and 5 is interesting because the integers are consecutive. I. e., we have 3 consecutive integers such that the sum of the squares of the first two is equal to the square of the third. In fact, this is the only set of consecutive positive integers for which this is true. Note that if negative integers are included, the consecutive set (-1, 0, 1) also satisfies the equation.

Carrying this a little further, we can say that the integers 3, 4, 5 and 6 form a cubic quartet in the sense that 3^{3} + 4^{3 }+ 5^{3} = 6^{3}. Thus, we have 4 consecutive integers such that the sum of the cubes of the first three is equal to the cube of the fourth. Very interesting! Almost like the previous example with the squares. Now the question becomes, "Can we extend this further?" I. e., are there n + 1 consecutive integers that satisfy a similar equation for higher powers like n = 4, 5, 6, etc. We can summarize this by asking whether there are integers n and m such that the equation

m^{n} + (m+1)^{n} + ... + (m+n-1)^{n }= (m+n)^{n}

has solutions for some integer values of m and n.

My conjecture is that there are no solutions when n > 3. I don't have proof for this but there are some simple cases that can be eliminated.

First, it is fairly easy to show that when n = 4, there are no solutions. This can be done by expanding the fourth order equation and show that the roots to this equation are not integers. I have worked this out through n = 8. After that it gets a little hairy.

Next, it is also easy to show there are no solutions when n = 5. This is done based solely on even or odd numbers. In fact this is going to be true whenever n mod 4 = 1. Thus for n = 5, 9, 13, ..., there are no solutions.

Beyond this, I haven't gone further.

E-mail me if you have any ideas about this!

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