Determining if $973$ is prime 
Without a calculator, determine if $973$ is prime or not

I was given this question to solve. I know $973$ is not prime. I was told a strategy to solve whether a number is prime or not is to test all the numbers less than the square root of $973$
So I would have to test till $32$
and i find $1,7,139$ and $973$ are factors of this number.  Basically, what I want to find out is are there any other strategies to solve this question? then i wouldn't have to check till 1-32 to see if any of the numbers are factors. 
 A: Another method is $973 = 1000-27$ which can be represented as $(10)^3-(3)^3$
Therefore, applying the identity that $(a)^3-(b)^3=(a-b)(a^2+b^2+ab)$, we see that
$973=(10)^3-(3)^3=(10-3)(100+9+30)=7\cdot139$
A: Not a clean method though but I used Fermat's factorization to find that,
$(31)^2<973<(32)^2$
Now applying the fact that a perfect square should end only in 0,1,4,5,6,9
, concentrate only on those numbers the difference of whose square and $973$ give these digits in the last place. Therefore concentrate only on those number whose last digit is either $2,3,7$ as the difference of squares of such numbers and 973 would end in numbers whose digits either end with $1$ or $9$.
Concentrating on such numbers and with a little bit of trial, we find that $(73)^2 - 973 = 5329 - 973 = 4356 = (66)^2$
Therefore, $973=(73-66)(73+66)=7\cdot
139$
Hence $973$ is not a prime.
A: For small numbers, there is no much better strategy than trying all prime divisors up to the square root.
Use divisibility criteria for the tiny divisors.


*

*$2$: check the last digit even;

*$3$: compute the sum of the digits (e.g. $975\to21\to3$);

*$5$: last digit must be $0$ or $5$;

*$7$: subtract twice the unit digits from the rest of the number (e.g. $973\to91\to7$);

*$11$: subtract the unit digits from the rest of the number (e.g. $2585\to253\to22$).
A: Alternatively, we can apply Fermat's Little Theorem where in if $p$ is a prime number and $a$ is any integer not divisible by $p$ then 
$a^{p-1} \equiv1 \mod p$
In this case, we can chose $a$ to be $2$ so as to keep things simple.
It's not hard to see that $(2)^{10}= 1024 \equiv51\mod973$
Therefore, on applying Fermat's Little Theorem
$(2^{10})^{97}$ . $2^2$ $\equiv(51)^{97}.2^2\mod 973$ $\not\equiv1\mod 973$
Hence, $973$ is not a prime number
A: Another way, $973 = 910+63$
This gives $973 = 7.(130+9) = 7 . 139$
A: If you're good at arithmetic, you can try this without a calculator $:)$
Wilson's theorem:
$p$ prime $\iff$ $(p-1)! \equiv -1 \pmod p$. 
Else, pocklington's test could be fast.
