Question 1209627
<pre>

{{{z^2 + 2z = -53 + 8i}}}

Let z = p + qi where p and q are integers.

{{{(p+qi)^2 + 2(p+qi) = -53 + 8i}}}

{{{p^2+2pqi+q^2i^2+2p+2qi=-53+8i}}}

{{{p^2+2pqi+q^2(-1)+2p+2qi=-53+8i}}}

{{{p^2+2pqi-q^2+2p+2qi=-53+8i}}}

Set the real parts equal on both sides and set
the imaginary parts on both sides equal

{{{system(p^2-q^2+2p=-53,2pqi+2qi=8i)}}}

Divide the second equation through by 2i

{{{system(p^2-q^2+2p=-53,pq+q=4)}}}

Solve the second equation for q

{{{pq+q=4}}}
{{{q(p+1)=4}}}
{{{q=4/(p+1)}}}

Substitute in

{{{p^2-q^2+2p=-53}}}
{{{p^2-(4/(p+1))^2+2p=-53}}}

By technology, there are two solutions, neither one integers

p = -0.446924168, which gives q = 7.232281305

and

p = -1.553075832, which gives q = -7.232281305

So, there are no such integer solutions, unless somebody can 
find a mistake that I've made.

Edwin</pre>