In solving equations, we first look into whether solutions exist or not. If they do, we analyze if the solutions are finite or infinite in number. If infinite, we give solutions in general form. With Diophantine equations, we search for solutions in positive integers.

Now my question is how to proceed as above for the following problems:

  1. $\displaystyle \frac{x^y - y^x}{x - y} = z^2$; I got one solution $(x, y, z) = (2, 3, 1)$.

  2. $x^3 + y^2 - x^2y + x^2y^3 - y^2x = x + y$; I got one solution $(x, y) = (1, 1)$.

The above solutions I got by trial-and-error. I would like to know the following:

  1. How to know if solutions exist or not for above equations?

  2. If they do, how to know if there are finitely or infinitely many?

  3. If finite how to find all solutions?

  4. If infinite how to characterize all solutions by a general formula?

  5. Is there any theory or particular method generalizing both the problems?

Thank you,

  • 2
    $\begingroup$ Non-linear (let alone exponential, like your first equation) diophantine equations are notoriously difficult: just look at the problem of determining that the Fermat equation has no nontrivial solutions for specific exponents, or Catalan's conjecture. $\endgroup$ Mar 12 '12 at 15:49
  • $\begingroup$ I am also interested to know the finding solutions of the first question. Whereas, the solution of second question is wrong. By inspection the solution given for first question is correct. But, how to get this solution without inspection by trail and error method? Is there any member, who had an idea for finding solutions for first problem. $\endgroup$
    – user26815
    Mar 13 '12 at 9:51

[This answer has been extensively edited to address the question more fully and correct minor errors.]

Diophantine equations are often very difficult. However, these two equations yield to fairly simple methods.

Equation 1

Since $z^2 > 0$, it is required that $(x^y – y^x) / (x – y) > 0$. To avoid division by zero, we must have $x \ne y$. Suppose that 0 < x < y.

Case 1: $x \geq 3$. Then $x^y > y^x$, and $(x^y – y^x) / (x – y) < 0$, so there are no solutions.

Case 2: x = 2. (x, y, z) = (2, 3, 1) is a solution, as stated in the question. If $y \geq 4$, then $2^y \geq y^2$ so there are no more solutions.

Case 3: x = 1. Then $(x^y – y^x) / (x – y) = (1 – y) / (1 – y) = 1$. Hence (x, y, z) = (1, n, 1) for any positive integer $n \geq 2$. This shows that the number of solutions is infinite. Note however that, as cases 2 and 3 illustrate, an infinite set of solutions to an equation does not necessarily contain all solutions.

For 0 < y < x, simply reverse the values of x and y in the above solutions (since $(x^y – y^x) / (x – y)$ is symmetrical in x and y).

A formula which gives all the solutions with x < y is (x, y, z) = (1+a, 1+b(1-a)+2a, 1) where a = 0 or 1 and b is any positive integer. However, this is a bit artificial.

Equation 2

This can be rewritten as:

$[x^3 - x] + [y^2 - y] + [x^2y^3 - x^2y - xy^2] = 0$

Then if x, y $\geq$ 2, each of the three terms enclosed by square brackets must be positive. To see this for the third term, note that if x, y $\geq$ 2:

$x^2y^3 \geq 2x^2y^2 > (x^2y + xy^2)$

Hence there are no solutions with $x, y \geq 2$. The only remaining possibilities for a solution in positive integers are x = 1 or y = 1. If x = 1 then:

$1 + y^2 - y + y^3 - y^2 = 1 + y$

$-y + y^3 = y$

$y(y^2 - 2) = 0$

Hence y = 0 or +/-$\surd2$ and so is not a positive integer. A similar argument shows that if y = 1 then x is not a positive integer. Hence equation 2 has no solution in positive integers (and questions 2-4 are not relevant to it).

Generalising from these problems

One lesson is to try simple approaches first. In the case of equation 1 the variables as exponents may suggest that it is very difficult. But if one asks simple questions the apparent difficulties can - in some cases - disappear. Key questions here are: Under what conditions is $(x^y - y^x)/(x-y) > 0$? What is the effect of putting x = 1? But there are very many diophantine equations for which such simple methods will not work.

  • $\begingroup$ ! I am agree with your approach. I am looking for first question. Can we do the first problem as per your method shown in second. Certainly no. But, if you can solve the first one, I am so thankful to you. $\endgroup$
    – user26950
    Mar 14 '12 at 9:23
  • $\begingroup$ Equation 1 looks much more difficult than equation 2, eg variables as exponents. As you say, my method for equation 2 would not work for equation 1. It's part of the fascination of diophantine equations that there are so many different types and so few methods that can be generalised. $\endgroup$ Mar 14 '12 at 10:31
  • $\begingroup$ Since posting the above comment I made some progress with equation 1 and have edited my answer. $\endgroup$ Mar 14 '12 at 11:41
  • $\begingroup$ Thank you for your response. Plz explain your work on first problem. $\endgroup$
    – user26950
    Mar 15 '12 at 4:24
  • $\begingroup$ Thank you so much for your beautiful explanation. $\endgroup$
    – user27021
    Mar 16 '12 at 4:22

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