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.

This is the outline of the proof found in the book "Proofs from THE BOOK" (P.17-22) by Aigner, Martin, Ziegler, Günter M. It is too long to put in a comment so I put it here. Please do not vote for it.

Lemma 1. For primes $p=4m+1$ the equation $s^2\equiv -1$ (mod $p$) has two solutions $s\in\{1,2,\dots,p-1\}$, for $p=2$ there is one such solution, while for primes of the form $4m+3$ there is no solution

Lemma 2. No number $n=4m+3$ is a sum of two squares.

Proposition. Every prime of the form $p=4m+1$ is a sum of two squares, that is, it can be written as $p=x^2+y^2$ for some natural numbers $x,y\in\mathbb{N}$

Theorem. A natural number $n$ can be represented as a sum of two squares if and only if every prime factor of the form $p=4m+3$ appears with an even exponent in the prime decomposition of $n$.

## Best Answer

$8290 \cdot 10 = (57^2 + 71^2)(10)$

$82900 = (57^2 + 71^2)(1^2 + 3^2)$

Rewrite using Brahmagupta–Fibonacci identity

$(57 + 213)^2+(171-71)^2$

$270^2 + 100^2 = 82900$