when you post question, please do not include spoilers and hints on titles demmy wrote: Let $A$ be set of 20 consecutive positive integers, Which sum and product of elements in $A$ not divisible by 23. Prove that product of elements in $A$ is not perfect square

mistakesinsolutions wrote: when you post question, please do not include spoilers and hints on titles demmy wrote: Let $A$ be set of 20 consecutive positive integers, Which sum and product of elements in $A$ not divisible by 23. Prove that product of elements in $A$ is not perfect square Sorry. I fixed it now

demmy wrote: Let $A$ be set of 20 consecutive positive integers, Which sum and product of elements in $A$ not divisible by 23. Prove that product of elements in $A$ is not perfect square Let $a,a+1,...,a+19$ be $20$ consecutive numbers.İf $a×(a+1)×(a+2)×...×(a+19)$ does not divisible by $23$ then $a,(a+1),..,(a+19)$ none is divisible by $23$.Then $a$ can be $23k+1,23k+2,23k+3$ 1)$a=23k+1$ $a×(a+1)×(a+2)×...×(a+19)$$\equiv$$20!$$(mod 23)$ $a×(a+1)×(a+2)×...×(a+19)$$\equiv$$p$$(mod 23)$ bu Wilson Theorem $2p$$\equiv$$22$$(mod 23)$ $\rightarrow$ $p$$\equiv$$11$$(mod 23)$ $11$ is NQR in modulo $23$ that is why it is imposible. 2)$a=23k+2$ then $23$$|$$(a+(a+1)+...+(a+19))$ 3)$a=23k+3$ $a×(a+1)×(a+2)×...×(a+19)$$\equiv$$p$$\equiv$$\frac{22!}{2}$$(mod 23)$ Bu Wilson Theorem $p$$\equiv$$11$$(mod 23)$ $11$ is NQR in modulo $23$ that is why it is imposible.

One link solution https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-19/issue-2/The-product-of-consecutive-integers-is-never-a-power/10.1215/ijm/1256050816.pdf