"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
"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
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 ?
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 ?
Hello, cedricc!
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.