//Check if there is a natural number i
//by which t can be divided without rest

//This code should only executed if it has been determined above that ((t - 1) % 3 == 0) is false

int matchFound=0;
int i=1;

while (i*i <= t) {			//Go through all natural numbers from 1 onwards
					//until i gets "too large"

	if (i*i == t) matchFound=1;	//If there is a natural number i that multiplied with itself is t,
					//t has an integer square root and is part of the function t=n²

	i++;				//Don't forget to actually count up!
	}

return matchFound;			//returns 1 if t is part of the function t=n², 0 if not