way to reduce time taken by this particular code for fibonacci numbers?

As a fun project, I started this code where I wanted to find the last digit of the Fibonacci number's squared each time. It works fine but I want to reduce the time taken by it... any suggestions?

trivia: I find the Fibonacci number, square them, and then find the sum.

import java.util.*;
import java.math.*;
public class FibonacciSumSquares {
private static BigInteger getFibonacciSumSquares(int n) {
if(n<=2)return BigInteger.valueOf(n);
BigInteger sum = BigInteger.valueOf(2);
long last = 1, lastTwo = 1, current = 0;
BigInteger lastBigInteger = BigInteger.ONE;
BigInteger lastTwoBigInteger = BigInteger.ONE;
BigInteger currentBigInteger;
boolean isUsePrimary = true;
for (int i = 3; i <=n; i++) {
if (isUsePrimary){
current = last + lastTwo;
current = current * current;
if (String.valueOf(current).length()<12) {
lastTwo = last;
last = current;
sum = sum.add(BigInteger.valueOf(current));
} else {
isUsePrimary = false;
lastTwoBigInteger = BigInteger.valueOf(lastTwo);
lastBigInteger = BigInteger.valueOf(last);
currentBigInteger = lastBigInteger.add(lastTwoBigInteger);
currentBigInteger = currentBigInteger.pow(2);
sum = sum.add(currentBigInteger);
}
} //end of outer if
else {
currentBigInteger = lastBigInteger.add(lastTwoBigInteger);
lastTwoBigInteger=lastBigInteger;
currentBigInteger = currentBigInteger.pow(2);
lastBigInteger= currentBigInteger;
sum = sum.add(currentBigInteger);
}
}//end of for
return sum.remainder(BigInteger.valueOf(10));
}//end of function
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
System.out.println(getFibonacciSumSquares(n));
}

}

Oh, you just want the last digit, that's a classic math twist

The catch is usually that if you only need the last digit, you can ignore all the other digits in all steps between, as long as it's only adding or multiplying.

Your formula is basically f(n+2) = [f(n) + f(n+1)]^2, and for that, the last digit is only dependent on the last digits of the previous results. Also, the last digit of a sum is only dependent on the last digits of every number you add.

So you can do mod 10 for every current or sum value, and you don't need BigInteger or even long to do that, you can just use int.

private static int getFibonacciSquaredSumMod10(int n) {
if (n <= 1)
return n;
int sum = 0;
int last = 1, lastTwo = 1, current = 0;
for (int i = 2; i <= n; i++) {
current = last + lastTwo;
current = (current * current) % 10;
lastTwo = last;
last = current;
sum = (sum + current) % 10;
}
return sum;
}