4

(i)

For a geometric sequence $a,ar,a{r}^2,a{r}^3$

$$a^2+(ar{)}^2+(a{r}^2{)}^2+(a{r}^3{)}^2=a^2(1+r^2+r^4+r^6)$$

We can immediately see that

$$1+r^2+r^4+r^6=r^6+2r^3+1+r^4-2r^3+r^2=(r^3+1{)}^2+(r^2-r{)}^2.$$

So for the original polynomial, we have

$$a^2((r^3+1{)}^2+(r^2-r{)}^2)=(a(r^3+1){)}^2+(a(r^2-r){)}^2.$$

Since we know that $a{r}^3$ and $a$ are integers, $a(r^3+1)=a{r}^3+a$ is also an integer.
And also we know that $a{r}^2$ and $ar$ are integers, $a{r}^2-ar$ is also an integer.
Hence, $a^2+b^2+c^2+d^2$ can be written as the sum of two perfect squares.

(ii)

Let the ratio be $r=2k$. For the geometric sequence $a,2ak,4a{k}^2,8a{k}^3,16a{k}^4$, we have

$$a^2+(2ak{)}^2+(4a{k}^2{)}^2+(8a{k}^3{)}^2+(16a{k}^4{)}^2=a^2(1+4k^2+16k^4+64k^6+256k^8)$$

We can separate

$$1+4k^2+16k^4+64k^6+256k^8$$

to

$$1+4k^2+4k^4+4k^4+4k^4+4k^4+64k^6+256k^8$$

Using our attention, we can immediately notice that it equals to

$$(1+4k^2+4k^4)+(4k^4)+(4k^4)+(4k^4+64k^6+256k^8)$$ $$=(16k^4+2k^2{)}^2+(2k^2{)}^2+(2k^2{)}^2+(2k^2+1{)}^2$$

Since they are all integers. Overall, the answers are

$$(\text{i})\boxed{{\left(a(r^3+1)\right)}^2+{\left(a(r^2-r)\right)}^2}$$ $$(\text{ii})\boxed{(a(16k^4+2k^2){)}^2+(2a{k}^2{)}^2+(2a{k}^2{)}^2+(a(2k^2+1){)}^2}$$