Let be a positive integer. Prove that there exist distinct positive integers such that (Source: 1997 IMO Shortlist N6)
The equation is known to have no nontrivial solutions for due to Fermat’s Last Theorem (if you’ve not read Fermat’s Enigma I highly recommend checking it out). This very similar looking equation, however, has an infinite number of solutions. In this blog post, we’ll examine how this difficult looking problem can be solved with one simple observation: .
For a concrete example of this, consider the case giving the equation Assume we multiply the equation by a number of the form . We then will have: Since we want the first part to be a perfect square, we should have and . Since the second part should be a perfect cube, we should have and . Finally, since the last part should be a perfect fourth power, and . Solving these system of linear congruences gives and ! We get back the equation , from which we read off the solution .
We now attempt to generalize this technique. We begin by multiplying by again to obtain We then want the exponents on the leftmost term to be divisible by , on the middle term to be divisible by , and on the rightmost term to be divisible by . This is equivalent to:
Notice that in the first case, the first and third conditions are equivalent to . Since , we can use the Chinese Remainder Theorem to know that has a unique solution mod .
On the other hand, for , in the case that is even, , therefore, we have to use an extended version of the Chinese Remainder Theorem. Thankfully, since in this case and , we are still guaranteed a unique solution for mod .
Substituting these solutions in for and gives a solution to the original equation. Once we have the congruences for and , we can then find an infinite parameterized version of solutions for .
For instance, in the case of above, we had . If we in general write and , we then find
This then translates into . Notice for , this gives the solution triple found above.
It’s pretty interesting to me how when you change one small thing in the formulation of the problem (subtract from the exponent on and add to the exponent of ) gives an entirely new problem, where we go from no solutions to an infinite number!