Proof that the product of 4 consecutive numbers is not a perfect square.

[speencer]

Hey,

I was thinking and I realized that this is true and I want to prove it but I have nowhere to start. If anyone knows any way to prove can you give me some advice on where to start.

 

[Opalg]

Hey,

I was thinking and I realized that this is true and I want to prove it but I have nowhere to start. If anyone knows any way to prove can you give me some advice on where to start.

Hint: Any four consecutive integers include one multiple of 4 and an odd multiple of 2.

 

[Evgeny.Makarov]

Hint: Any four consecutive integers include one multiple of 4 and an odd multiple of 2.

And obviously "an odd multiple of 2" here means the product of 2 and an odd number, not a multiple of 2 that is odd.

 

[chisigma]

Hey,

I was thinking and I realized that this is true and I want to prove it but I have nowhere to start. If anyone knows any way to prove can you give me some advice on where to start.

The correct formulation should be...

Prove that the product of four consecutive numbers all different from 0 is not a perfect square...

Kind regards

$\chi$ $\sigma$

 

[soroban]

Hello, speencer!

Prove that the product of four consecutive positive integers is not a perfect square.

The four consecutive positive integers are: .[tex]x,\,x+1,\,x+2,\,x+3[/tex]

Suppose their product is a perfect square.
. . [tex]x(x+1)(x+2)(x+3) \:=\:k^2\;\text{ for some integer }k.[/tex]

We have: .. . . [tex]x(x+3)\cdot(x+1)(x+2) \:=\:k^2[/tex]

. . . . . . . . . . . . [tex](x^2+3x)(x^2+3x+2) \:=\: k^2[/tex]

. [tex]\big[(x^2+3x+1)-1\big]\big[(x^2+3x+1) + 1\big] \:=\:k^2[/tex]

. . . . . . . . . . . . . . . [tex](x^2+3x+1)^2 - 1^2 \:=\:k^2[/tex]

And we have: .[tex](x^2+3x+1)^2 - k^2 \:=\:1[/tex]
. . The difference of two squares is 1.

The only case is when: [tex]x^2+3x+1 \:=\:1\,\text{ and }\,k\:=\:0[/tex]

If [tex]k = 0[/tex], then one of the four integers must be zero.
We have our contradiction.

Therefore, the product of four consecutive positive integers cannot be a square.

 

[caffeinemachine]

Hey,

I was thinking and I realized that this is true and I want to prove it but I have nowhere to start. If anyone knows any way to prove can you give me some advice on where to start.

To know how to start its a good idea to "get your hands dirty". Start putting values of $n$ in $f(n)=n(n+1)(n+2)(n+3)$.

You get:
$f(1)=24, f(2)=120, f(3)=360, f(4)=840$.
Each of these is equal to a one less a square.
So one can guess that $f(n)$ always is equal to $k^2-1$ for some $k$.
Then one can go ahead in the direction of proving it which many have done in the previous posts.