Proof, that product of 5 numbers cannot be square of a positive interger

**cedricc** · April 2nd 2010, 02:48 AM

"Prove that the product of five consecutive numbers cannot be the square of a positive integer."

I have tried using examples, and it shows that this is true, but how do i prove this? Please show working/reasoning if possible.

Thanks

**Bacterius** · April 2nd 2010, 03:07 AM

Hello,
easy enough. Given five consecutive numbers $a$ , $b$ , $c$ , $d$ and $e$ , for their product to be a square they need to combine all their prime factors with even powers. For instance, taking only two numbers for simplicity, for their product to be a square we need for instance $7 \times 7 = 7^2$ or $7^4 \times 3^2 = (49 \times 3)^2$ , and $7 \times 3 = 7^1 \times 3^1$ is not a square because every prime factor doesn't have an even power.

Now assume some prime number greater than three divides $a$ . It obviously doesn't divide $a + 1 = b$ , $a + 2 = c$ , $a + 3 = d$ , $a + 4 = e$ . Therefore, this prime factor is only present once in the product $abcde$ , hence has an odd exponent (or power) ( $1$ ) and thus $abcde$ is not a square.

Now just validate the statement with the two prime numbers smaller or equal to three, namely 2 and 3 (just take $a = 2$ and $a = 3$ , since any multiple is suitable), to conclude the proof. Oh, and also validate for $a = 1$ and $a = 0$ ! And for negative numbers, you only have to check up to $a = -4$ , because the product of five negative numbers is a negative number which cannot be a square !

Does it make sense ?

**Bacterius** · April 2nd 2010, 03:41 AM

I just noticed a flaw in my proof. I'll correct it here : Note that if $p^2$ , $p$ being a prime number greater than three, divides $a$ , then $abcde$ has $p$ with an even power. However, if $a = p^2$ with $p$ prime and $p > 3$ , then neither $a + 1, a + 2, a + 3, a + 4$ can be squares, applying the same reasoning and introducing the linear increase of the difference between two consecutive squares.

Does ... does it make sense ?

**Soroban** · April 2nd 2010, 06:14 AM

Hello, cedricc!

Prove that the product of five consecutive integers
cannot be the square of a positive integer.

Let the five consecutive integers be: . $n-2,\:n-1,\:n,\:n+1,\:n+2$

Their product is: . $P \:=\:(n-2)(n-1)n(n+1)(n+2) \;=\;n(n^2-1)(n^2-4) \;=\;n(n^4-5n+4)$

Assume $P$ is a square.

Since $P$ contains the factor $n$ , it must contain the factor $n^2$ .
. . That is, $n^4-5n^2+4$ has a factor of $n$ . . . and we can factor it out.

Hence: . $P \;=\;n^2\left(n^3-5n + \frac{4}{n}\right)$

Since $P$ is an integer, the cubic expression must be an integer.

. . Then $\frac{4}{n}$ must be an integer . . . $n$ is a factor of 4.

Hence: . $n \:=\:\pm1,\:\pm2,\:\pm4$

And we have: . $\begin{array}{|c|c|} n & P \\ \hline \pm1 & 0 \\ \pm2 & 0 \\ \pm4 & \pm720 \end{array}$ . None of these is the square of a positive integer.

. . Q.E.D.

**ICanFly** · April 2nd 2010, 06:56 AM

2 Soroban
Imho this statement: "Since

is an integer, the cubic expression must be an integer" is wrong. For instans, P=11, ${n^2} = 9<br />$ then
"cubic expression" must be $\frac{{11}}{9}<br />$

**awkward** · April 3rd 2010, 01:17 PM

Originally Posted by Soroban

Assume $P$ is a square.

Since $P$ contains the factor $n$ , it must contain the factor $n^2$ .

This isn't true. For example, consider the perfect square 36. It is divisible by 18, but not by 18^2.

If n is prime, that is a different matter.

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