Question 554273
When faced with a similarly impossible problem that is not answered here, you could give the artofproblemsolving website community a try. The frightfully brainy members of those forums would find an answer if there is one. They used to have a lot of this kind of problem.
MY ANSWER
The two solutions are 11000 and 57464.
THE LONG CASE WORK  (Sorry, it's the best I could do).
Let A be the first digit, B the second digit, C the third and fifth digits, and D the fourth digit.
The number would be ABCDC.
We know that A is not zero (or else it would not be a 5 digit number), and that:
{{{A-C=B-D}}}
{{{CD=A^2-1}}}
{{{B-A=sqrt(C)}}}
The last condition means that C can only be 0, 1, 4, or 9, or else
{{{B-A=sqrt(C)}}} would not be an integer.
It also tells you that {{{B>=A}}} to make {{{B-A=sqrt(C)>=0}}}, and since
{{{A-C=B-D}}} --> {{{D-C=B-A}}}, it also means that {{{D-C>=0}}} and {{{D>=C}}} 
WITH C=0,
{{{CD=0=A^2-1}}} so A=1
Also {{{B-A=sqrt(C)=0}}}, so A=B=1
Then {{{A-C=B-D}}} turns into {{{1-0=1-D}}}, so D=0.
That would give you the number 11000.
There could be other answers with C=1, C=4, or C=9
{{{CD=A^2-1}}} means that the product CD can only be 0, 3, 8, 15, 24, 35, 48, 63, or 80, for A=1, 2, 3, ...9, respectively. However those values have C (1, 4, or 9) as a factor.
WITH C=1
as a factor, CD=D, and we are limited to 3 of the 9 possibilities above:
CD=D=0 (for A=1), CD=D=3 (for A=2), and CD=D=8 (for A=3).
D=0 does not meet the requirement that {{{D>=C}}}
For the other cases, we can calculate B from {{{B-A=sqrt(C)}}} --> {{{B=A+sqrt(C)}}}
and see if {{{A-C=B-D}}} is true
For A=2 and D=3, {{{B=A+sqrt(C)=2+sqrt(1)=3}}}
{{{A-C=2-1=1}}} and {{{B-D=3-3=2}}} It does not work.
For A=3 and D=8, {{{B=A+sqrt(C)=3+sqrt(1)=4}}}
{{{A-C=3-1=2}}} and {{{B-D=4-8=-4}}} It does not work.
WITH C=4 or C=9
The choices for value of the product CD listed above narrow to
CD=8={{{4*2}}} (for C=4, A=3) with D=2,
CD=24={{{4*6}}} (for C=4, A=5) with D=6,
CD=63={{{9*7}}} (for C=9, A=8) with D=7
The first choice (C=4, A=3, D=2) and the last one (C=9, A=8, D=7) are eliminated because they do not comply with {{{D>=C}}}.
For the remaining possibility, we calculate B from {{{B-A=sqrt(C)}}} --> {{{B=A+sqrt(C)}}}
and see if {{{A-C=B-D}}} is true.
For C=4, A=5, D=6 --> {{{B=A+sqrt(C)=5+sqrt(4)=5+2=7}}}
Then {{{A-C=5-4=1}}} and {{{B-D=7-6=1}}} We have another solution (57464).