Skip to main content.

Solution to puzzle 105: Difference of nth powers

Let x, y, n be positive integers, with n > 1.  How many solutions are there to the equation xn − yn = 2100?


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

Case n = 2

We have x2 − y2 = (x − y)(x + y) = 2100.
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 2100, 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 = 2a
x + y = 2b

Solving these simultaneous equations, we obtain

x = 2b−1 + 2a−1
y = 2b−1 − 2a−1

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

Therefore there are 49 solutions to the equation x2 − y2 = 2100.

Case n = 4

x4 − y4 = 2100 implies y4 + (225)4 = x4, 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.)

Case even n > 4

We will show that there are no solutions with even n > 4 to the more general equation xn − yn = 2k, 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 xn − yn = 2k, and n is minimal.
Letting n = 2m, we have (xm − ym)(xm + ym) = 2k.
But then xm − ym = 2a, for some integer a greater than or equal to 0, with m > 2, contradicting the minimality of n.

Therefore no solution exists for even n > 4.

Case odd n

We will show that there are no solutions with odd n to the more general equation xn − yn = 2k, where k is a positive integer.

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

Clearly x, y are either both even or both odd.  Since
(x − y)(xn−1 + xn−2y + ... + yn−1) = 2k,
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
un − vn = 2k−n.
If k − n > 0, this contradicts the minimality of k.
If k − n = 0, we have no solution in positive integers to un − vn = 1.

Therefore no solution exists for odd n.

Conclusion

Putting the results together, the equation xn − yn = 2100 has 49 solutions in positive integers x, y, n, with n > 1.


Further reading

  1. Irrationality by Infinite Descent
  2. Fermat's Infinite Descent
  3. Infinite Descent versus Induction

Source: Original

Back to top