NOT a maximal ideal of the ring of Gaussian integers

Question

I have to show that $I= \{ a+bi : a,b \in \mathbb{Z} , a=5m , b=5n\}$ is NOT a maximal ideal of $\mathbb Z[i]$.

For this if I take $J= \{ a+bi : 5\ $divides $ a^2+b^2 \}$ then , I can have that $4+3i \in J$ but NOT in $I$, and $1+5i \in \mathbb Z[i]$ but NOT in $J$. Then $ I \subset J\subset \mathbb Z[i]$. But the problem lies in showing J as a subring:

for $x, y\in J$, say $ x = a+ib$ and $y = c+id$, $x-y = (a-c)+i(b-d)$. Then $5$ must divide $(a-c)^2 + (b-d)^2$. Now 5 divides $a^2+b^2$ and $c^2+d^2$, but how do I know that 5 divides $2(ac-bd)$ ?

Edited to add another related query: For an earlier problem of similar nature, where ideal I = $\{a+ib:$ a and b are multiples of 2} , J was assumed to be $\{a+ib:$ 2 divides $\ a^2+b^2$}. I understand that here that $\ a^2+b^2$=$(a+ib)(a−ib)$. Is there any similar structure to get non maximal ideals of Z[i]?

Any help would be duly acknowledged.

Answer 1

The non-trivial ideal $(2+i)\mathbb Z[i]$ contains $I=5\mathbb Z[i]$, owing to the factorisation $5=(2+i)(2-i)$. This settles the matter.