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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
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1 Answer 1

up vote 4 down vote accepted

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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