Hello Amir, some similar problems were posted before.Paul Erdos proved that the product of several consecutive integers is not a perfect power

Hi. Could you please post a solution ?

If $p|x_3,p>2$, then $p\not|x_1x_2x_4x_5\to v_p(x_3)$ is even. Therefore $x_3=n^2$ or $x_3=2n^2$. If $p>3$ and $p|x_2$ then $v_p(x_2)$ is even. It give $x_3-1=x_2=m^2,2m^2,3m^2,6m^2$. By analogy $x_3+1=x_4=l^2,2l^2,3l^2,6l^2$. By analogy $x_1=k^2,2k^2,3k^2,6k^2,x_5=q^2,2q^2,3q^2.6q^2$. Because we have only four form we have $i<j,x_i=cx^2,x_j=cy^2\to 4\ge x_j-x_i|\ge c(y^2-x^2)=c(y+x)(y-x)\to c=1,x=1,y=2$. It is easy to chek, that $1*2*3*4*5$ is not square.

Hi! This and many other problems with product of consecutive natural numbers is on Russian TST. Solution of this problem: grater common divisor of any two numbers in left side is $1, 2, 3$ or $6$ => if product is square then all of numbers $a, a+1, a+2, a+3, a+4$ equals $n^2, 2n^2, 3n^2$ or $6n^2$. But we have $4$ types and $5$ numbers => exist numbers of one type ($a, b$)=> $4 \geq a-b \geq 5$ for $a \geq 2$ => for $a \geq 2$ $a(a+1)(a+2)(a+3)(a+4)$ is not square => $QED$ (case $a=1$ trivial)

Attachments:
1-1ru2.pdf (176kb)

Hi,I downloaded this but the language is not English.

I write, that problem at Russian