I will list the properties of a row of Pascal triangle that will be used in the test.
1-Every number of a given row $n$ is divisible by the row number $n$ if the row number is a prime except of course the first and last number $1$.
2-The sum of all the elements of any given row with row number n, including the first and last $1$, is $2^n$. This result is valid for a row number $n$ prime or composite. This means that we do not need to add all the elements of a row to know the total value of the sum. We can just write it down for a given row with a given row number $n$.
Test: The test consists in evaluating the following quantity: $T=(2^n -2)/n$. If the result is an integer then we can conclude that $n$ is a prime. If the result is not an integer, we can conclude that $n$ is a composite number. This test bypasses the calculation and summation of the binomial coefficients to get the sum of all the elements of a given row $n$.
Examples:
1-For row number $7$, we have $T=(2^7-2)/7=126/7=18$ and integer so $7$ is a prime
2-For row number $15$, we have $T=(2^{15}-2)/15=32766/15=2184.40$ so $15$ is not a prime.
Can this result be proven?
I can imagine a situation where, for large numbers, we run into $2$ or more elements of a given row $n$, initially not individually divisible by $n$ adding up to a sum that is divisible by $n$. I don't know enough to prove this will never happen for large numbers.
Is this a case of "too good to be true"?
Best Answer
Nice observation, but not quite. It is indeed too good to be true.