linguistic:paraphrasing
Equivalent statements of the theorem

There are several alternative ways to state Fermat's Last Theorem that are mathematically equivalent to the original statement of the problem.

In order to state them, we use the following notations: let N be the set of natural numbers 1, 2, 3,..., let Z be the set of integers 0, ±1, ±2,..., and let Q be the set of rational numbers a / b, where a and b are in Z with b ≠ 0. In what follows we will call a solution to x + y = z where one or more of x, y, or z is zero a trivial solution. A solution where all three are nonzero will be called a non-trivial solution.

For comparison's sake we start with the original formulation.

Most popular treatments of the subject state it this way. It is also commonly stated over Z:

The equivalence is clear if n is even. If n is odd and all three of x, y, z are negative, then we can replace x, y, z with − x, − y, − z to obtain a solution in N. If two of them are negative, it must be x and z or y and z. If x, z are negative and y is positive, then we can rearrange to get (− z) + y = (− x) resulting in a solution in N; the other case is dealt with analogously. Now if just one is negative, it must be x or y. If x is negative, and y and z are positive, then it can be rearranged to get (− x) + z = y again resulting in a solution in N; if y is negative, the result follows symmetrically. Thus in all cases a nontrivial solution in Z would also mean a solution exists in N, the original formulation of the problem.

This is because the exponents of x, y, and z are equal (to n), so if there is a solution in Q, then it can be multiplied through by an appropriate common denominator to get a solution in Z, and hence in N.

A non-trivial solution a, b, c ∈ Z to x + y = z yields the non-trivial solution a / c, b / c ∈ Q for v + w = 1. Conversely, a solution a / b, c / d ∈ Q to v + w = 1 yields the non-trivial solution ad, cb, bd for x + y = z.

This last formulation is particularly fruitful, because it reduces the problem from a problem about surfaces in three dimensions to a problem about curves in two dimensions. Furthermore, it allows working over the field Q, rather than over the ring Z; fields exhibit more structure than rings, which allows for deeper analysis of their elements.

Examining this elliptic curve with Ribet's theorem shows that it does not have a modular form. However, the proof by Andrew Wiles proves that any equation of the form y = x (x − a)(x + b) does have a modular form. Any non-trivial solution to x + y = z (with p an odd prime) would therefore create a contradiction, which in turn proves that no non-trivial solutions exist.

In other words, any solution that could contradict Fermat's Last Theorem could also be used to contradict the modularity theorem. So if the modularity theorem were found to be true, then it would follow that no contradiction to Fermat's Last Theorem could exist either. As described above, the discovery of this equivalent statement was crucial to the eventual solution of Fermat's Last Theorem, as it provided a means by which it could be "attacked" for all numbers at once.