Let $p_{1}>p_{2}>p_{3}>p_{4}>p_{5}$ are prime numbers smaller than natural $x$
Find the smallest ( in respect to $x$) solution of the system of equations: $$\begin{cases} 2x-p_{4}=p_{5}^{a_{1}}\\2x-p_{2}=p_{4}^{a_{2}}\\2x-p_{1}=p_{4}^{a_{3}}p_{5}^{a_{4}}\\2x-p_{3}=p_{1}^{a_{5}}\\2x-p_{5}=p_{3}^{a_{6}}\end{cases}$$ Where $a_{i}$ is natural number for each $i$
So far i tried to write down c++ program but it does not work.
Thanks for help.
Not a full solution:
First note, you can order by magnitude on the left hand side: $$\begin{cases}2x-p_5=p_3^{a_6}\\2x-p_4=p_5^{a_1}\\2x-p_3=p_1^{a_5}\\2x-p_2=p_4^{a_2}\\2x-p_1=p_4^{a_3}p_5^{a_4}\end{cases}$$
Second note, all left hand sides are greater than x, this means every right hand side has at least 1 exponent greater than 1.
Third note, no p values are 2 by distributivity of multiplication over subtraction, and our strict ordering.
Fourth note, Subtracting left hand sides bottom to top gives us differences of values (potentially minimum if guessed correctly)
From all this we can gather the following
$$\begin{eqnarray}x>p_1>p_2>p_3>p_4>p_5>2\\2p_5\mid p_4+p_3^{a_6}\\2p_4\mid p_1-p_2\\2p_3\mid p_1^{a_5}+p_5\\2p_1\mid p_4^{a_3}p_5^{a_4}+p_3\\2p_5\mid p_1-p_4\\2p_5\mid p_1+p_3^{a_6}\\2p_4\mid p_2+p_5^{a_1}\\2p_4\mid p_1+p_5^{a_1}\\\end{eqnarray}$$
There are more restrictions, but no solution to my knowledge.