# Whether a prime number $p$ can be written in the form $3A + 2B$, where $A,B \in \mathbb{N}$.

I would like to know whether or not a prime number $p$ can be written in the form $$p = 3A + 2B,$$ where $A$ and $B$ are positive integers.

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If you require both $A$ and $B$ to be positive, then this is false for $P = 2$ and $P = 3$, and trivial for $P \geq 5$: let $A = 1$. – Andrew Uzzell Jan 3 '13 at 10:16

This is an application of the Frobenius Coin Problem, whose solution is given by the following theorem.

Theorem Suppose that $m,n \in \mathbb{N}$ satisfy $\gcd(m,n) = 1$. Then $mn - m - n$ is the largest integer that cannot be written as $ma + nb$, where $a,b \in \mathbb{N}_{0}$.

Observe that $3A + 2B = 3(A - 1) + 2(B - 1) + 5$. As $\gcd(3,2) = 1$, the theorem yields the following statements:

• Any integer $> 3 \cdot 2 - 3 - 2 = 1$ can be written as $3a + 2b$, where $a,b \in \mathbb{N}_{0}$.

• Any integer $> 1$ can thus be written as $3(A - 1) + 2(B - 1)$, where $A,B \in \mathbb{N}$.

• Therefore, any integer $> 1 + 5 = 6$ can be written as $3A + 2B$, where $A,B \in \mathbb{N}$.

Notice that the numbers $1$, $2$, $3$, $4$ and $6$ cannot be put in the required form, whereas $5 = 3 \cdot 1 + 2 \cdot 1$ can. Therefore, the answer to your problem is ‘all prime numbers $\geq 5$’.

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