"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

- April 2nd 2010, 03:48 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 - April 2nd 2010, 04:07 AMBacterius
Hello,

easy enough. Given five consecutive numbers , , , and , 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 or , and*is not a square because every prime factor doesn't have an even power*.

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

Does it make sense ? - April 2nd 2010, 04:41 AMBacterius
I just noticed a flaw in my proof. I'll correct it here : Note that if , being a prime number greater than three, divides , then has with an even power. However, if with prime and , then neither 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) - April 2nd 2010, 07:14 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: .

Their product is: .

Assume is a square.

Since contains the factor , it must contain the factor .

. . That is, has a factor of . . . and we can factor it out.

Hence: .

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

. . Then must be an integer . . . is a factor of 4.

Hence: .

And we have: . . None of these is the square of a positive integer.

. . Q.E.D.

- April 2nd 2010, 07:56 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, then

"cubic expression" must be - April 3rd 2010, 02:17 PMawkward