# Product of 5 consecutive integers cannot be a perfect square [duplicate]

Problem: Show that the product of any five consecutive positive integers cannot be a perfect square.

Proof: Let $N=n \times (n+1) \times (n+2) \times (n+3) \times (n+4)$, where $n \in \mathbb{N}$. We know the that N must have at least:

1. one factor is divisible by 5,
2. two or three factors are even,
3. one factor of the even factors, is divisible by $4$,
4. and one factor is divisible by $3$.

To proof the statement above we need to assume that the product of any five consecutive positive integers is a perfect square to lead us to a contradiction. Hence, I know there will be cases for us to test for. However, I am unsure where to start after mentioning everything above. Can anyone help me from here?

• Proof is stated here: mathforum.org/library/drmath/view/65589.html – Kal S. Dec 28 '13 at 23:28
• There isn't another way to prove it? – Username Unknown Dec 28 '13 at 23:33
• This question has already been asked in math stackexchange and the link was the only answer that was given. I don't think that an alternative proof will differ a lot from this one. math.stackexchange.com/questions/304417/… – Kal S. Dec 28 '13 at 23:35
• That is why I asked it because I find the answer to be to vague and tedious – Username Unknown Dec 28 '13 at 23:37
• The motivation in the OP's last comment (1) should have been mentioned in the post, (2) contradicts what is mentioned in the post, (3) is quite gracious to the author of mathforum's answer. – Did Dec 28 '13 at 23:56

Consider the product $n(n+1)(n+2)(n+3)(n+4)$. A perfect square $k$ will have a factorization $k=a_1^{2m_1}a_2^{2m_2}...a_n^{2m_n}$ , i.e., every factor will appear an even number of times. But this will not happen for $n(n+1)(n+2)(n+3)(n+4)$; there will always be a factor that appears an odd number of times. There are two main cases here:

1) The middle term $(n+2)$ is not a perfect square , and,

2)The middle term is a perfect square .

Case 1): If $(n+2)$ is not a perfect square, the product, then, cannot be a perfect square, because the factors of the middle term $(n+2)$ will appear an odd number of times, unless $(n+2)$ is itself a perfect square, since $$2(n+2)>(n+4); g(\frac{(n+2)}{2}) < n$$, where $g(\frac{n+2}{2})$ is the greatest integer less than $(n+2/2)$.

Basically, any factor $a_i$ of $(n+2)$ will be between $ka_i$ and $(k+1)a_i$ for some positive integer $k$. This is because of the fact that $$ka_i < n(n+1)(n+2)(n+3)(n+4)< (k+1)a_i$$ when $n+2$ is not a perfect square.

If $(n+2)$ is itself a perfect square, then neither $(n+1)$ , nor $(n+3)$ are perfect squares, and you can repeat a similar argument.

• This is wrong. For instance, if $n+2$ is even, then $2$ is a common factor of $n+2$, $n$, and $n+4$, and so you can't be sure (without more work) that $n+2$ has to be a square. – Eric Wofsey May 25 '16 at 1:50
• What are you contradicting? I don't quite understand. You $can$ be sure that the highest power of 2 in $n+2$ is an even power (hence 1, or 4, or 16, or...), which is all that is necessary, because the same argument is then applied to all other possible prime factors of $n+2$. – Martin Kochanski Jun 2 '16 at 14:15

We need to think about prime factorisations.

Assume the product is a perfect square.

It is intuitively clear that for any prime p, greater than (or equal to) 5, p is a factor of at most one of the integers. We also know that for p greater than (or equal to) 5, p must have even power, because we're square rooting and getting a rational.

This leaves us to consider primes 2 and 3. These are the only primes which can appear in the prime decomposition of our integers with an odd power. Now we need to split into cases.

Integers can be of 4 forms:

power(2)=even, power(3)=even implies m^(2)

power(2)=odd, power(3)=even implies 2m^(2)

power(2)=even, power(3)=odd implies 3m^(2)

power(2)=odd, power(3)=odd implies 6m^(2)

We have 5 integers total, so two must be of the same form.

Supposing we have 2 of the form m^(2) means we must have 1 and 4 in our list. Checking this, we get (1)(2)(3)(4)(5)=120 which isn't a square.

Supposing we have 2 of the form 2m^(2) means we have two squares with a difference of 1 or 2, which clearly doesn't happen.

Supposing we have 2 of the form 3m^(2) gives another contradiction, as for this to happen, two squares need to have a difference of 1, which is false.

Supposing we have 2 of the form 6m^(2), we get two multiples of 6 but with a difference of only 4.

In all cases, we find contradictions.