Le Monde puzzle [#1028]

Back to standard Le Monde mathematical puzzles (no further competition!), with this arithmetic one:

While n! cannot be a squared integer for n>1, does there exist 1<n<28 such that 28(n!) is a square integer? Does there exist 1<n,m<28 such that 28(n!)(m!) is a square integer? And what is the largest group of distinct integers between 2 and 27 such that the product of 28! by their factorials is a square?

The fact that n! cannot be a square follows from the occurrence of single prime numbers in the resulting prime number decomposition. When considering 28!, there are several single prime numbers like 17, 19, and 23, which means n is at least 23, but then the last prime in the decomposition of 28! being 7 means this prime remains alone in a product by any n! when n<28. However, to keep up with the R resolution tradition, I started by representing all integers between 2 and 28 in terms of their prime decomposition:

primz=c(2,3,5,7,11,13,17,19,23)
dcmpz=matrix(0,28,9)
for (i in 2:28){
 for (j in 1:9){
    k=i
    while (k%%primz[j]==0){ 
      k=k%/%primz[j];dcmpz[i,j]=dcmpz[i,j]+1}}
}

since the prime number factorisation of the factorials n! follows by cumulated sums (over the rows) of dcmpz, after which checking for one term products

fctorz=apply(dcmpz,2,cumsum)
for (i in 23:28)
  if (max((fctorz[28,]+fctorz[i,])%%2)==0) print(i)

and two term products

for (i in 2:28)
for (j in i:27)
 if (max((fctorz[28,]+fctorz[i,]+fctorz[j,])%%2)==0) 
  print(c(i,j))

is easy and produces i=28 [no solution!] in the first case and (i,j)=(10,27) in the second case. For the final question,  adding up to twelve terms together still produced solutions so I opted for the opposite end by removing one term at a time and

for (a in 2:28)
  if (max(apply(fctorz[-a,],2,sum)%%2)==0) print(a)

exhibited a solution for a=14. Meaning that

2! 3! …. 13! 15! …. 28!

is a square.

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