can anyone show me the proof that "the product of four positive integers in arithmetic progression cannot be the square of an integer".

thanks!

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- July 19th 2006, 12:40 AMsherylproof please...
can anyone show me the proof that "the product of four positive integers in arithmetic progression cannot be the square of an integer".

thanks! - July 19th 2006, 05:17 AMSoroban
Hello, sheryl!

I know a proof for a similar problem, but I've hit a wall.

Let me show you the groundwork . . . maybe you can finish it.

Quote:

The product of four positive integers in arithmetic progression

cannot be the square of an integer.

Let the four numbers be: .

Suppose their product**is**a square.

Then we have: .

Regroup: .

Multiply: .

And we have: .

Multiply: .

And this is*supposed*to lead to a contradiction . . . but I can't find it.

Anyone? .Anyone?

- July 19th 2006, 11:01 AMSoroban
Hello, all!

The problem I referred to is:

"Prove that the product of four consecutive positive integers cannot be a square."

Let the integers be:

Suppose their product**is**a square.

Then we have: .

Regrouping: .

We have: .

. . then: .

Hence: .

We have: "The difference of two squares is "

But this is true for only.

Then: .

. . and is**not**a positive integer.

- July 19th 2006, 07:56 PMmalaygoelQuote:

Originally Posted by**Soroban**

Let gcd(a,d)=c

Let a=cm, d=cn

Then your expression will be c^4 m(m+n)(m+2n)(m+3n)

Now, since m and n are relatively prime, any two factors are relatively prime to each other(may have common factor of 2 or 3)

Hence their product cannot be a cannot be a perfect square.

Keep Smiling

Malay