Question 1070474
There are {{{9}}} positive digits.
Of those, 2, 3,5,and 7 are prime.
The other 5 are 1,4,6,8, and 9.
 
Since even and odd digits alternate,
and the 5 digits must be
even digits 4, 6, and 8,
and odd digits 1 and 9,
the even digits must be A, C, and E,
while B and D are 1 and 9.
Since A+D=E, with E being a digit, D cannot be 9,
so it has to be {{{highlight(D=1)}}}, and {{{highlight(B=9)}}} .
Then {{{A+D=B}}} means {{{A+1=9}}} --> {{{A=9-1}}} --> {{{highlight(A=8)}}} .
 
Of the 9 available digits exactly 3 are square,
{{{1=1^2}}} ,
{{{4=2^2}}} ,
{{{9=3^2}}} .
Those will be the middle digits, B, C, and D.
Since{{{B=9}}} and {{{D=1}}} , it must be {{{highlight(C=4)}}} .
By elimination,
{{{highlight(E=6)}}} .
The 5-digit password is
{{{highlight(ABCDE=89416)}}}