The Product Meets The Square

Let $a,b,c$ denote three consecutive integers (in that order) such that their product $abc$ is a perfect square. Since they are consecutive, we have $a=b-1$ and $c=b+1,$ so that $b(b^2-1) = (b-1)(b)(b+1) = abc$

Now, since $\gcd(b,b^2-1) = 1$ and their product is a square, we must have that $b^2 - 1 = n^2$ is a perfect square, or written another way, $(b-n)(b+n) = b^2 - n^2 = 1.$ Since the only way to write $1$ as the product of two integers is $1 \times 1 = 1 = -1 \times -1,$ it follows that $2b = (b-n)+(b+n) = 2\times (\pm 1) = \pm 2 \implies b = \pm 1 \implies b^2 - 1 = 0$

However, then $abc = b(b^2-1) = 0$ so that at least one of $a,b,c$ is zero, showing that $\boxed{\text{no,}}$ there aren't any three consecutive positive integers such that their product is a perfect square.

Write $b=n$ , such that $a=n-1$ and $c=n+1$ . Then,

$n(n-1)(n+1)=m^2$

for some integer $m$ . Combine the factors on the left to obtain:

$n^3-n=m^2$

or

$x^3-x=y^2$

an elliptic curve. A bit of tinkering with a computer algebra program then shows that it has no integral solutions. Thus, there are no integers $a, b, c,$ and $m$ such that

$abc=m^2$

Let the consecutive integers be x - 1, x , and x + 1. Then x(x^2 -1) = y^2 Let r = gcd(x,y), so that x = ra, y = rb with (a,b) = 1. Substituting and dividing by r, a(r^2 a^2 - 1) = rb^2. Since (a,b) = 1, this implies that a|r. Let r = am. Substituting and dividing by a, a^4 m^2 - 1 = mb^2. Then m = 1, and a^4 - 1 = b^2, or a^4 - b^2 = 1, which is impossible, Since the difference of 2 positive squares cannot equal 1. Ed Gray