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Solution to puzzle 65: Consecutive integer products

Skip restatement of puzzle.Show that each of the following equations has no solution in integers x > 0, y > 0, n > 1.

  1. x(x + 1) = yn
  2. x(x + 1)(x + 2) = yn

Both results may be proved in a similar way.

1.  x(x + 1) = yn

Let p be a prime factor of y, occurring k times in its factorization.  Then p occurs kn times in the prime factorization of yn.
By the Fundamental Theorem of Arithmetic, the prime factorization of yn is unique, and is the same as that of x(x + 1).
Hence pkn occurs in x(x + 1).

The greatest common divisor (gcd) of x and x + 1 is 1.  This follows from the fact that any divisor of two numbers must also divide their difference.
Hence pkn occurs in x, or in x + 1, but not in both.
The same is true for each of the prime factors of y, and therefore x = an, x + 1 = bn, for some positive integers a, b.
We then have bn − an = 1, which is impossible if a and b are positive integers.

Therefore x(x + 1) = yn has no solution in integers x > 0, y > 0, n > 1.

2.  x(x + 1)(x + 2) = yn

Set w = x + 1, so that (w − 1)w(w + 1) = w(w2 − 1) = yn.
Since gcd(w, w2 − 1) = 1, we have w = an, w2 − 1 = bn, for some positive integers a, b.
We then have (a2)n − bn = 1, which is impossible if a2 and b are positive integers.

Therefore x(x + 1)(x + 2) = yn has no solution in integers x > 0, y > 0, n > 1.

Remark

A generalization of these results, that the product of any number of consecutive positive integers is never a perfect power, was proved by Erdös and Selfridge.  See the references below.


Further reading

  1. Product of consecutive numbers as a power
  2. Paul Erdös
  3. Information about Paul Erdös
  4. How to discover a proof of the fundamental theorem of arithmetic

Source: Traditional

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