# Solution to puzzle 65: Consecutive integer products

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

- x(x + 1) = y
^{n}
- x(x + 1)(x + 2) = y
^{n}

Both results may be proved in a similar way.

## 1. x(x + 1) = y^{n}

Let p be a prime factor of y, occurring k times in its factorization. Then p occurs kn times in the prime factorization of y^{n}.

By the Fundamental Theorem of Arithmetic, the prime factorization of y^{n} is unique, and is the same as that of x(x + 1).

Hence p^{kn} 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 p^{kn} 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 = a^{n}, x + 1 = b^{n}, for some positive integers a, b.

We then have b^{n} − a^{n} = 1, which is impossible if a and b are positive integers.

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

## 2. x(x + 1)(x + 2) = y^{n}

Set w = x + 1, so that (w − 1)w(w + 1) = w(w^{2} − 1) = y^{n}.

Since gcd(w, w^{2} − 1) = 1, we have w = a^{n}, w^{2} − 1 = b^{n}, for some positive integers a, b.

We then have (a^{2})^{n} − b^{n} = 1, which is impossible if a^{2} and b are positive integers.

Therefore x(x + 1)(x + 2) = y^{n} 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

- Product of consecutive numbers as a power
- Paul Erdös
- Information about Paul Erdös
- How to discover a proof of the fundamental theorem of arithmetic

Source: Traditional

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