## Simple, Yet Rewarding, Diophantine Equation

**Point of post:** Often in mathematics it is enjoyable to solve simple problems in fields outside of one’s comfort zone, especially one’s that occur with uncanny frequency in real life (and by real life I mean when people studying undergraduate algebra ask you how to solve them haha). The one I have in mind is solving the basic Diophantine Equation

**The Problem**

In this first part of the post we will give several proofs that the only solution to is .

**Proof (1): **We first see that it is sufficient to solve this for since is a solution if and only if is a solution. So note that

implies that and . Thus, we have that is equal to either , , , or and similarly for . So, we consider the table

where the column represents the values of and the horizontal the values of . So considering that we see that the only possibilities are

Where we’ve included the first two since there is no reason that can’t be or . But, if either of these are true we see that , and so we’ll consider the solution to later. But, we also see that both the third and fourth set of equations imply that since adding the equations in both systems gives

since the only even prime is . Lastly, we see that the last two are impossible since in both cases adding the equations in the system gives

which is impossible. Thus, on all counts we see that it must be true that . So, the only equation of the form with possible solutions is . To solve this we notice once again that the equation implies that and so that and . So, drawing up a multiplication table again gives us that

Thus, the only possibilities are

The first has the trivial solution and the last two are impossible since addition of the equations in each system gives

Thus, after all of this work we have conclude that the only prime for which an equation of the form has solutions is and the solutions to are . Thus, it follows that the solutions of are

**Proof (2): **We first note that in the following table holds

So that is even precisely when and have the same parity. So, we consider equation when and are both even, or both odd separately.

First suppose that and then equation may be rewritten as

But, this clearly implies that so that . Now, suppose that and then equation may be written as

which also implies that so that . It follows that if has a solution then . Thus, solving reduces to solving . But, this was done in the last proof so that we may conclude that has precisely the solutions . .

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