• July 19th 2006, 12:40 AM
sheryl
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 AM
Soroban
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: . $a,\;a+d,\;a+2d,\;a+3d$

Suppose their product is a square.
Then we have: . $a(a+d)(a+2d)(a+3d)\;=\;k^2$

Regroup: . $[a(a+3d)]\,[(a+d)(a+2d)] \;=\;k^2$

Multiply: . $[a^2+3ad)\,(a^2+3ad + 2d^2)\;=\;k^2$

And we have: . $[(a^2 + 3ad + d^2) - d^2]\,[(a^2+3ad + d^2) + d^2] \;= \;k^2$

Multiply: . $(a^2+3ad + d^2)^2 - d^4 \;= \;k^2$

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

Anyone? .Anyone?

• July 19th 2006, 11:01 AM
Soroban
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: $a,\;a+1,\;a+2,\;a+3$

Suppose their product is a square.
Then we have: . $a(a+1)(a+2)(a+3) \:=\:k^2$

Regrouping: . $\left[a(a+3)\right]\,\left[(a+1)(a+2)\right] \:=\:k^2$

We have: . $[a^2 + 3a]\,[a^2+3a +2] \:=\:k^2$

. . then: . $\left[(a^2 + 3a + 1) - 1\right]\,\keft[(a^2 + 3a + 1) + 1\right] \:= \:k^2$

Hence: . $(a^2 + 3a + 1)^2 - 1^2\:=\:k^2\quad\Rightarrow\quad (a^2 + 3a + 1)^2 - k^2\:=\:1$

We have: "The difference of two squares is $1.$"

But this is true for $(\pm1)^2 - 0^2$ only.

Then: . $a^2 + 3a + 1 \:= \:\pm1\quad\Rightarrow\quad\left\{\begin{array}{cc }a(a +3) \:= \:0 \\ (a+1)(a+2) \:= \:0\end{array}$

. . and $a$ is not a positive integer.

• July 19th 2006, 07:56 PM
malaygoel
Quote:

Originally Posted by Soroban
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.

[size=3]
Let the four numbers be: . $a,\;a+d,\;a+2d,\;a+3d$

Suppose their product is a square.
Then we have: . $a(a+d)(a+2d)(a+3d)\;=\;k^2$

It is just enthusiasm you talked about/

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