$n = p_1p_2p_3p_4$, prove that if $n \lt 1995$ then $d_9-d_8 \not = 22$

Question

For each integer $n = p_1p_2p_3p_4$, where $p_1,p_2,p_3,p_4$ are distinct primes.
Let $(d_1=1) \lt d_2 \lt d_3 \lt \cdots \lt d_{15}\lt (d_{16} = n)$ be $16$ positive divisors of $n$.
Prove that if $n \lt 1995$ then $d_9-d_8 \not = 22$.

My Work
Without loss of generality assume $p_1 \lt p_2 \lt p_3 \lt p_4$. Then it is easy to see that $d_9-d_8 = p_2p_3 - p_1p_4$
Now how to continue from here?

Source: This is a problem from 1995 Irish Mathematical Olympiad.

Answer 1

If $d_9-d_8=p_2p_3-p_4p_1$, then suppose it's equal to 22 then clearly p cannot be 2 and since $n<1995$ it follows that $p_4<19$.
Now if you suppose that 3 is not on of primes then you get $n>1995$, same happens for 5 and 7 so $p_1=3,p_2=5,p_3=7$ and now you are left only with $p_4=11,13,17$, so check cases.