"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

- Apr 2nd 2010, 02:48 AMcedriccProof, that product of 5 numbers cannot be square of a positive interger
"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 - Apr 2nd 2010, 03:07 AMBacterius
Hello,

easy enough. Given five consecutive numbers $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, $\displaystyle d$ and $\displaystyle e$, 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 $\displaystyle 7 \times 7 = 7^2$ or $\displaystyle 7^4 \times 3^2 = (49 \times 3)^2$, and $\displaystyle 7 \times 3 = 7^1 \times 3^1$*is not a square because every prime factor doesn't have an even power*.

Now assume some prime number**greater than three**divides $\displaystyle a$. It obviously doesn't divide $\displaystyle a + 1 = b$, $\displaystyle a + 2 = c$, $\displaystyle a + 3 = d$, $\displaystyle a + 4 = e$. Therefore, this prime factor is only present__once__in the product $\displaystyle abcde$, hence has an*odd exponent*(or power) ($\displaystyle 1$) and thus $\displaystyle abcde$ 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 $\displaystyle a = 2$ and $\displaystyle a = 3$, since any multiple is suitable), to conclude the proof. Oh, and also validate for $\displaystyle a = 1$ and $\displaystyle a = 0$ ! And for negative numbers, you only have to check up to $\displaystyle a = -4$, because the product of five negative numbers is a negative number which cannot be a square !

Does it make sense ? - Apr 2nd 2010, 03:41 AMBacterius
I just noticed a flaw in my proof. I'll correct it here : Note that if $\displaystyle p^2$, $\displaystyle p$ being a prime number greater than three, divides $\displaystyle a$, then $\displaystyle abcde$ has $\displaystyle p$ with an even power. However, if $\displaystyle a = p^2$ with $\displaystyle p$ prime and $\displaystyle p > 3$, then neither $\displaystyle a + 1, a + 2, a + 3, a + 4$ can be squares, applying the same reasoning and introducing the linear increase of the difference between two consecutive squares.

Does ... does it make sense ? (Worried) - Apr 2nd 2010, 06:14 AMSoroban
Hello, cedricc!

Quote:

Prove that the product of five consecutive integers

cannot be the square of a positive integer.

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

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

Assume $\displaystyle P$ is a square.

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

. . That is, $\displaystyle n^4-5n^2+4$ has a factor of $\displaystyle n$ . . . and we can factor it out.

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

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

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

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

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

. . Q.E.D.

- Apr 2nd 2010, 06:56 AMICanFly
2 Soroban

Imho this statement: "Since http://www.mathhelpforum.com/math-he...9a0fdaaa-1.gif is an integer, the cubic expression must be an integer" is wrong. For instans, P=11, $\displaystyle {n^2} = 9

$ then

"cubic expression" must be $\displaystyle \frac{{11}}{9}

$ - Apr 3rd 2010, 01:17 PMawkward