Prove that the product of four consecutive positive integers is not a power of any positive integer with exponent greater than or equal to 2.

Question

For the case of the product of two or three consecutive integers, this is easy to prove (using coprimality and expressing 1 in terms of the factors quickly leads to a contradiction). However, for the product of four consecutive positive integers, I’m unable to make further progress. Although there is a related paper by Erdős that confirms this result, I believe there must be a more concise and elementary proof, since this is Exercise 1.6.3 in Róbert Freud’s textbook on number theory

Answer 1

Let $P(n) = n(n+1)(n+2)(n+3)$. First, observe the identity:

$$ (n^2 + 3n + 1)^2 = n^4 + 6n^3 + 11n^2 + 6n + 1 = P(n) + 1, $$

which means:

$$ P(n) = (n^2 + 3n + 1)^2 - 1. $$

Suppose for the sake of a contradiction that $P(n) = y^m$ for some integer $y \ge 1$ and exponent $m \ge 2$. Then:

$$ X^2 - y^m = 1, $$

where $X = n^2 + 3n + 1$.

By Mihăilescu's Theorem, the only solution in the positive integers to:

$$ x^a - y^b = 1 \quad \text{with} \quad x, y > 0,\, a, b \ge 2 $$

is:

$$ 3^2 - 2^3 = 1. $$

So the only possibility is:

$$ X = 3, \quad y = 2, \quad m = 3. $$

But then:

$$ n^2 + 3n + 1 = 3 \Rightarrow n^2 + 3n - 2 = 0, $$

which has discriminant:

$$ \Delta = 9 + 8 = 17, $$

so the roots are:

$$ n = \frac{-3 \pm \sqrt{17}}{2}, $$

which are not integers.

Therefore, $P(n)$ cannot be a perfect power for any positive integer $n$. □

Answer 2

Suppose there exist $n,m\in\mathbb{Z_{>0}},l\in\mathbb{Z_{>1}}$ such that

$$n(n+1)(n+2)(n+3) = m^l$$

If $n\equiv{1,5}\pmod{6}$, $n$ is coprime to $6$, and thus coprime to $(n+1)(n+2)(n+3)$ .Therefore, there exists $a\in\mathbb{Z_{>0}}$ such that $(n+1)(n+2)(n+3) = a^l$, which is impossible. See this post.

Similarly, if $n\equiv{2,4}\pmod{6}$, then $n+3$ is coprime to $n(n+1)(n+2)$, and there will exist $b\in\mathbb{Z_{>0}}$ such that $(n+1)(n+2)(n+3) = b^l$, leading to a contradiction as well.

If $n\equiv{0}\pmod{6}$, then $n+1$ is coprime to $n(n+2)(n+3)$, and there are $c,d\in\mathbb{Z_{>0}}$ such that

$$n+1=c^l, n(n+2)(n+3) = d^l$$

The case $l=2$ is ruled out by Guruprasad's comment. When $l>2$, the following two inequalities can cause contradiction.

$$d^l = n(n+2)(n+3) > {(n+1)}^{3} = c^{3l}$$

$$d^l = n(n+2)(n+3)<(n+1)^3+2(n+1)^2 = c^{3l}+2c^{2l} < c^{3l} + lc^{3(l-1)} < {(c^3+1)}^l$$

Finally, if $n\equiv{3}\pmod{6}$, similarly there will exist $f,g\in\mathbb{Z_{>0}}$ which are coprime to each other and satisfying

$$n+2 = f^l, n(n+1)(n+3) = g^l$$

When $l\geqslant{4}$, since $f>1$, it follows that

$$g^l = n(n+1)(n+3) < (n+2)^3 = f^{3l}$$

$$g^l = n(n+1)(n+3) > f^{3l} - 2f^{2l} - f^{l} \geqslant {f^{3l} - f^{3l-3} - f^{3l-6}} > (f^3-1)^l$$

When $l=3$,

$$g^3 = n(n+1)(n+3) < (n+2)^3 = f^9$$

$$g^3 = n(n+1)(n+3) > f^{9} - 2f^{6} - f^{3} > (f^3-1)^3$$

In both cases we have $f^3-1 < g < f^3$, which is a contradiction. $l=2$ is impossible as well. Therefore, the product of four consecutive positive integers could not be a perfect power.