A quick way, say in a minute, to deduce whether $1037$ is a prime number So with $1037 = 17 \cdot 61$, is there a fast method to deduce that it's not a prime number?
Say $1037 = 10^3+6^2+1$.  Does $a^3 + b^2 + 1$ factorize in some way?
As part of their interviews, a company is asking whether a number is prime. I have never studied number theory, and I am not aware of a strategy for this apart from polynomial factorization.
I am guessing that for a number they give you, it would have to not be prime, as the only way to see that a number is a prime by hand is to test all primes below $\sqrt{N}$.

$u\mid v$ is read as "$u$ divides $v$" and $u\nmid v$ is read as "$u$ does not divide $v$."

Obviously $2\nmid 1037$ since it has an odd last digit.
By the "Divisibility by $3$ Rule," it follows that $3\nmid 1037$ since $3\nmid 1+0+3+7=11$.
Obviously $5\nmid 1037$ since the last digit is not $0$ nor $5$.
If $7\mid 1037$ then $7\mid 1037-7=1030$ but $1030=103\times 10$, and $103$ is prime and $7\nmid 10$.
If $11\mid 1037$ then $11\mid 1037-77=960$ but $960=96\times 10$ and $11\mid 99$ so $11\nmid 96$ since it is off by a remainder of $3$ (and obviously $11\nmid 10$).
Try for $13$ this time, and then for $17$. You will see that it works for $17$, which indicates that $17\mid 1037$ and $1037$ is not prime.

You can try this method for all numbers, and you only need to try the first prime numbers less than the square root of the number you are testing with; for example, say you don't know if $61$ is a prime number. Then, apply this method.
Now, $64=8^2$ so $\sqrt{61}$ is pretty close to $8$. And since $7^2=49<61$, then all you need to do is see if $2,3,5$ or $7$ divide $61$. If they don't, then $61$ is prime (which it is).
This method does not have an official name, so it might as well be called a "trial divisibility check" by primes. Thanks to users that commented below who corrected me as I thought this method was the following.
A well known method is the "Sieve of Eratosthenes."

Tip:
A good number of primes to remember off the top of your head is all of the primes up to $127$. This has helped me, and $127$ is a pretty special number. There are many good properties about $127$; for example, $127$ is a Mersenne prime (a prime of the form $2^p-1$ for prime $p$), and it is the $31$st prime number, $31$ also being a Mersenne prime.
If you want to remember more prime numbers after $127$, you can. (I did all the way up to $383$, so it is possible.)

Apologies if this answer is primarily opinion-based.
