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- Feb 7, 2012

- 2,702

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

- Jan 30, 2012

- 2,492

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

- Feb 13, 2012

- 1,704

The correct formulation should be...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.

Kind regards

$\chi$ $\sigma$

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

The four consecutive

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 can

- Mar 10, 2012

- 834

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

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.