On prime factors of composite integers

Question

I have a question on prime factors of composite integers.

It is known that for any composite integer N, N must contain a prime factor $p$ $\leqslant$ $\sqrt n$. I then thought of the following question:

Suppose N $\leqslant$ $x$ $\leqslant$ N + 3, and all integers in that range are composite. Is it true that for at least one $x$ in that range, $x$ contains exactly one prime factor $p$ $>$ $\sqrt n$ ? Could this be proved for all sets of 4 consecutive, composite integers?

I tried a few examples, and noticed that this pattern holds (obviously, a few examples won't cut it - a formal proof is needed to verify its truth, or a counterexample for its falsity). For example, let us take the set of 4 composite, consecutive integers {32, 33, 34, 35}. Assume that N = 32. The square root of 32 is around 5.65. So we know that 32 must have a prime factor less than 5.65. Indeed, it does, since 32 = 2^5. However, there does exist at least one integer in that set that contains exactly one prime factor larger than 5.65, namely, 33, which is 3 x 11, and 11 $>$ 5.65. The numbers 34 and 35 also meet this criterion, since both 17 and 7 are greater than 5.65.

I do know (and it is obvious) that if such an integer M existed with one prime factor $p_i$ $>$ $\sqrt n$, then that integer M must contain exactly one, and only one, prime factor of that sort, since if another prime factor $p_j$ existed in the prime factorization of M, and $p_j$ $>$ $\sqrt n$, then $p_i$$p_j$ $>$ M, which contradicts the existence of $p_j$ in the prime factorization of M itself. However, the existence of at least one integer, $x$, in the given range N $\leqslant$ $x$ $\leqslant$ N + 3, and where x contains only one prime factor $p$ $>$ $\sqrt n$ - I am not sure of. But I am confident that it could be true.

So is it possible that this is the case for every set of 4 consecutive, composite integers? Could it formally be proven?

Answer 1

well, no

smallest

1518      23    = 2 * 3 * 11 * 23
1519      31    = 7^2 * 31
1520      19    = 2^4 * 5 * 19
1521      13    = 3^2 * 13^2

another

7137      61    = 3^2 * 13 * 61
7138      83    = 2 * 43 * 83
7139      59    = 11^2 * 59
7140      17    = 2^2 * 3 * 5 * 7 * 17

FIVE

5828      47    = 2^2 * 31 * 47
5829      67    = 3 * 29 * 67
5830      53    = 2 * 5 * 11 * 53
5831      17    = 7^3 * 17
5832       3    = 2^3 * 3^6 

SIX

   28032      73    = 2^7 * 3 * 73
   28033      97    = 17^2 * 97
   28034     131    = 2 * 107 * 131
   28035      89    = 3^2 * 5 * 7 * 89
   28036     163    = 2^2 * 43 * 163
   28037      53    = 23^2 * 53

EIGHT

  290783     271    = 29 * 37 * 271
  290784     233    = 2^5 * 3 * 13 * 233
  290785     311    = 5 * 11 * 17 * 311
  290786     419    = 2 * 347 * 419
  290787     227    = 3 * 7 * 61 * 227
  290788     523    = 2^2 * 139 * 523
  290789     269    = 23 * 47 * 269
  290790     359    = 2 * 3^4 * 5 * 359