Possible last digits of any square number :$$0,1,4,5,6,9$$
Possible combinations that can give perfect squares which end in $7$ are:$$(1,6)$$
Find the numbers whose squares are ending with $1,6$ and less than $2017$.
List of such numbers whose squares end in $1$ are : $1,9,11,19,21,29,31,39,41$. Check if any of the squares of these numbers when subtracted from $2017$, gives you a perfect square.
$$2017 - 1^2 = 2017 - 1 = 2016$$
$$2017 - 9^2 = 2017 - 81 = 1936 = 44^2$$
$$2017 - 11^2 = 2017 - 121 = 1896$$
$$....$$
$$2017 - 41^2 = 2017 - 1681 = 336$$
Find all the possible pairs, one such pair being $(9,44)$.After going through all the above mentioned cases, you can eventually conclude that $(9,44)$ is the only pair which satisfies the given condition.
Leading off with the following.
If $g$ is a primitive element of a finite field then multiplication by $g$ takes a non-zero square to a non-zero non-square and vice versa.
So if we can represent $0$ as a sum of two non-squares, then multiplication by $g$ shows that $0$ can also be written as a sum of two squares:
$$
0=a^2+b^2.
$$
This implies that
$-b^2=a^2$ and a fortiori that $-1=(a/b)^2$ is a square. This is known to be the case in $\Bbb{F}_q$ if and only if $q\equiv1\pmod4$.
Consequently:
There exists arbitrarily large finite fields such that the element $0$ cannot be written as a sum of two non-squares of that field. More precisely, this happens in the field $\Bbb{F}_q$ whenever $q\equiv-1\pmod4$.
A more interesting result is that any non-zero element $z$ of a finite field $\Bbb{F}_q$, $q$ and odd number $>5$, can be written as a sum of two non-squares. This can be seen as follows.
Assume first that $z$ is a square. Then $g^{-1}z$ is a non-square. By the well- known result we can write it as a sum of two squares
$$
g^{-1}z=x^2+y^2.
$$
Because $g^{-1}z$ is a non-square, we can deduce that $x$ and $y$ must both be non-zero. This means that the elements $gx^2,gy^2$ are both non-squares, and
$$
z=gx^2+gy^2
$$
is a presentation of the required type.
If $z=ga^2$ is a non-square then we need the result (see e.g. Ireland and Rosen) that the equation
$$
x^2+y^2=1\qquad(*)
$$
has $q-\eta(-1)$ solutions (here $\eta$ is the unique multiplicative character of order two, so equal to the Legendre symbol in the case of a prime field). The equation $(*)$ is equivalent to
$$
a^2x^2+a^2y^2=a^2,
$$
so the equation $x^2+y^2=a^2$, too, has $q-\eta(-1)\ge q-1$ solutions. At most $4$ of those solutions have either $x=0$ or $y=0$. So if $q>5$, then we are guaranteed the existence of elements $x\neq0\neq y$ such that
$g^{-1}z=a^2=x^2+y^2$. Again, it follows that
$$
z=gx^2+gy^2
$$
is a presentation of $z$ as a sum of two non-square.
The OP noted themself that in the fields of $3$ or $5$ elements there are too few non-squares. For example in $\Bbb{F}_5$ the only non-squares are $2$ and $3$, and we cannot write either of those as sums of two non-squares.
For the case of the prime fields the elegant solution by Mikhail Ivanov is surely better than this argument.
Best Answer
Adapting Mikhail Ivanov's argument to a slightly different but AFAICT equivalent question to fit here. Some of the elements appeared also in my answer to that question.
Every non-zero element of $\Bbb{Z}/p\Bbb{Z}$ is either a square or a non-square If $a=b^2$ is a non-zero square, then, as $p>5$ we have $$ a=b^2=(3b/5)^2+(4b/5)^2 $$ as a sum of two non-zero squares.
On the other hand, if $a$ is a non-zero non-square then $ab^2$ is a non-zero non-square for any $b\neq0$. Furthermore, we get all the non-squares in this way. If $a=x^2+y^2$ with $xy\neq0$, then $ab^2=(bx)^2+(by)^2$, so it suffices to show that we can write at least one of the non-squares as a sum of two non-zero squares. Let's pretend for one time's sake that $\Bbb{Z}/p\Bbb{Z}$ has an "order", so $0<1<2<\ldots<p-1$. Let $a$ be the smallest non-square in this order. Clearly $a>1$. It follows that $a-1=b^2$ is a square, where $b\neq0$. This implies that $a=1+b^2$ is a sum of two non-zero squares. Therefore so are all the other non-squares.