Let x, y, n be positive integers, with n > 1. How many solutions are there to the equation x^{n} − y^{n} = 2^{100}?

We consider separately the cases n = 2, n = 4, even n > 4, odd n.

We have x^{2} − y^{2} = (x − y)(x + y) = 2^{100}.

By the Fundamental Theorem of Arithmetic, (x − y) and (x + y) must be powers of 2.

Further, since (x − y) and (x + y) differ by an even number, they must be either both even or both odd.

Since their product equals 2^{100}, they must both be even, and so (x − y) > 1.

Hence, since y > 0, we must have, for integers a and b, with 0 < a < b and a + b = 100

x − y = 2^{a}

x + y = 2^{b}

Solving these simultaneous equations, we obtain

x = 2^{b−1} + 2^{a−1}

y = 2^{b−1} − 2^{a−1}

Hence, for the solutions we seek, (a, b) = (1, 99), (2, 98), ... , (49, 51).

Therefore there are 49 solutions to the equation x^{2} − y^{2} = 2^{100}.

x^{4} − y^{4} = 2^{100} y^{4} + (2^{25})^{4} = x^{4}, which by Fermat's Last Theorem has no solution. (Fermat's Last Theorem for exponent 4 was first proved by Fermat himself, using his method of infinite descent.)

We will show that there are no solutions with even n > 4 to the more general equation x^{n} − y^{n} = 2^{k}, where k is a non-negative integer.

We assume that a solution exists, and choose positive integers x, y, n (with n > 2) and non-negative integer k such that x^{n} − y^{n} = 2^{k}, and n is minimal.

Letting n = 2m, we have (x^{m} − y^{m})(x^{m} + y^{m}) = 2^{k}.

But then x^{m} − y^{m} = 2^{a}, for some integer a 0, with m > 2, contradicting the minimality of n.

Therefore no solution exists for even n > 4.

We will show that there are no solutions with odd n to the more general equation x^{n} − y^{n} = 2^{k}, where k is a positive integer.

We assume that a solution exists, and choose positive integers x, y, n, k such that x^{n} − y^{n} = 2^{k}, and k is minimal.

Clearly x, y are either both even or both odd. Since

(x − y)(x^{n−1} + x^{n−2}y + ... + y^{n−1}) = 2^{k},

if x, y are odd, the second term contains an odd number of odd terms, and so is odd, which is impossible.

Hence, x, y are even. Now set x = 2u, y = 2v, so that

u^{n} − v^{n} = 2^{k−n}.

If k − n > 0, this contradicts the minimality of k.

If k − n = 0, we have no solution in positive integers to u^{n} − v^{n} = 1.

Therefore no solution exists for odd n.

Putting the results together, the equation x^{n} − y^{n} = 2^{100} has 49 solutions in positive integers x, y, n, with n > 1.

Source: Original