Proof that the product of 7 successive positive integers is not a square. Can someone give me a hint for this problem (for (8-9) grade student):Proof that the product of 7 successive positive integers is not a square.
(I've found a proof for general case is given in : here)
 A: Assume that the product $n\cdot(n+1)\cdot\ldots(n+6)$ is a square number $m^2$. 
Then we proceed with the following steps:


*

*Whenever one of the factors is divisible by a prime number $p\geq7$, that prime must occur with an even power in that number.

*The sequence $n,\ldots,n+6$ contains a subsequence $u,u+2,u+4$ of odd numbers.

*Exactly one of $u,u+2,u+4$ is divisible by 3, and at most one of $u,u+2,u+4$ is divisible by 5.

*As a consequence of (1), each element $x$ of $\{u,u+2,u+4\}$ is either


*

*a square (in which case $x\equiv1\pmod8$),

*$3$ times a square (in which case $x\equiv3\pmod8$),

*$5$ times a square (in which case $x\equiv5\pmod8$), or

*$15$ times a square (in which case $x\equiv7\pmod8$).


*As a consequence of (2), the only possibility is $u=a^2$, $u+2=3b^3$ and $u+4=5c^2$.

*Now $u+3$ is neither divisible by $3$ nor by $5$, so $u+3$ is the product of a power of $2$ and an odd square (by (1)).

*Since $u\equiv1\pmod8$, we have $u+3\equiv4\pmod8$, which can only happen if $u+3$ is disivible by $4$ but not by $8$.

*(6) and (7) imply that $u+3$ is itself a square. Since it has distance $3$ to $u=a^2$, this implies that $u=1$ and thus $m^2 = 1\cdot\ldots\cdot7 = 5040$, a contradiction since $5040$ is not a square.
