Elements of the sequence have a prime factors an element of the sequence I am reading the following problem:

For the sequence $T=3, 7, 11, 15, 19, 23, 27 ...$ prove that every
number in $T$ has a prime factor that is also in $T$

My approach:
The sequence is of the form $4\cdot n + 3$
Each of the numbers in the sequence has a unique decomposition of prime factors $2^a\cdot 3^b \cdot 5^b \cdot 7^d ...$
We can express the prime factors as follows:
$(4\cdot n + 2)^a \cdot (4\cdot n + 1)^b \cdot (4\cdot n + 3)^c$ for $n \ge 0$
If we assume that we do not have any prime factor that is also part of $T$ then we would have prime factors of the form:
$(4\cdot n + 2)^a \cdot (4\cdot n + 1)^b$
For the simple case that $a = 1 \space b = 1$ we have:
$(4\cdot n + 2) \cdot (4\cdot n + 1) = 16 \cdot n^2 + 4 \cdot n + 8 \cdot n + 2 = 4n(4n + 3) + 2 = 4k + 2$ (where $k = 4\cdot n+ 3)$
Which means that we can not get a number in $T$ without having a prime factor of the form $(4\cdot n + 3)^c$
Update based on the comments of @Peter and @Asher2211:
The prime factors can be only of the form $(4\cdot n + 1)$ or $(4\cdot n + 3)$
If we assume that we can have a number in $T$ with prime factors not in $T$ we would have:
$x = (4\cdot n + 1)^a = 1 + a\cdot (4n) + \frac{a(a-1)}{2!}\cdot (4n)^2 +... = 1 + 4nk\space$ where $k = a + \frac{a(a-1)}{2!}\cdot (4n)....$
hence we can not get a number that is in $T$
 A: An experimental approach:
We rewrite expression as:
$4n+3=3(n+1)+n$
$[n, (n+1)]=1$
So if 3 divides n we can have numbers which are divisible by 3 in the sequence. These numbers make following progression:
$15, 27, 39=3\times 13, 51=3\times 17, \cdot\cdot\cdot 12k+3$
That is the terms of this progression have at least one prime other than 3 as a factor.
Now suppose 3 does not divide n, so we may consider following cases:
1): $\begin {cases}n=3m+1\\n=3m-2\end {cases}or:  n\equiv 1\bmod 3\equiv -2\bmod 3$
which gives this progression:
$19, 31, 43, 55=5\times 11, \cdot\cdot\cdot 12k_1-5$
The terms of these progression are a multiple of 5 or have at least a prime other than 3 and 5.
2): $\begin {cases}n=3m-1\\n=3m+2\end {cases}or: n\equiv -1\bmod 3\equiv 2\bmod 3$
which gives this progression:
$23, 35=5\times 7, 47, 59, 71, 83, 95, \cdot\cdot\cdot 12k_2-1$
similar to case 1 the terms of this progression are a multiple of 5 or have at least a prime other than 3 and 5.
We can see that the factors of $4n+3$ cover all primes and we can conclude that any term of resulting sequence from $4n+3$ contains a prime already exist in the factors of previous terms.
