Discovering All Positive Integer Pairs a, b for Which the Expression is Prime

Finding all positive integer pairs ( (a, b) ) such that the expression ( frac{b}{4} sqrt{frac{2a-b}{2ab}} ) is a prime number can be broken down into several steps. Let's explore this step-by-step, starting with the initial conditions and moving towards the final solution.

Initial Expression and Conditions

We begin with the expression: [ frac{b}{4} sqrt{frac{2a-b}{2ab}} text{ is prime.} ] To simplify, we first investigate under what conditions this expression can equal a prime number. We start by rewriting the inner square root part as follows: [ sqrt{frac{2a-b}{2ab}} frac{p}{q} text{ with } p, q in mathbb{N} text{ and } text{gcd}(p, q) 1. ] This leads to the equation: [ 2a - b p^2 quad text{and} quad 2ab q^2. ] By rearranging, we get: [ 2aq^2 - p^2 bp^2q^2. ] We see that ( b eq 0 ) because if ( b 0 ), the expression is undefined. Additionally, since we need the expression to be a prime number, ( p ) and ( q ) must be coprime.

Introducing a Positive Integer c

We can multiply both sides of the equations by a positive integer ( c ), which doesn't affect the root but extends the options for ( frac{b}{4} ). Therefore, we have the following system of equations: [ 2a - b cp^2 quad text{and} quad 2ab cq^2. ] Solving for ( a ) and ( b ), we get: [ a frac{cq^2}{p^2} quad text{and} quad b frac{cq^2 - p^2}{2}. ]

Conditions on p and q

We need to consider the parity of ( p ) and ( q ): - If ( p ) and ( q ) are both even, this violates the condition ( text{gcd}(p, q) 1 ). So, at least one of them must be odd. - If ( p ) and ( q ) have opposite parity, we need ( 4c ) to make ( a ) an integer. - If both ( p ) and ( q ) are odd, we need ( 2c ) to make ( a ) an integer. Substituting these values into our initial expression, we need to ensure that the following expression is prime: [ frac{cq^2 - p^2}{8} cdot frac{p}{q}. ] Rewriting, we get: [ frac{cq - pq}{8q}. ] For this expression to be prime, the denominator ( 8q ) must be a factor of the numerator ( cq - pq ). This means ( q ) must be a factor of the term ( cq - pq ). Therefore, ( qc ) must be considered. Additionally, since ( p ) and ( q ) have opposite parity, the product ( pq ) is odd, implying that both ( p ) and ( q ) are odd.

Special Cases and Solutions

Let's analyze the special cases step-by-step: - **Case 1: p is even** If ( p ) is even, it can only be divisible by 2 to prevent an extra prime factor. We need ( 4c ). Setting ( p 2 ) and ( q 3 ) and finding ( c 4 ), we get the solution ( (a, b) (39, 30) ) with the resulting prime ( 5 ). - **Case 2: p is odd, q is even** If ( p ) is odd and ( q ) is even, we need ( p 1 ) and ( q 2 ). This gives ( c 16 ) and the solution ( (a, b) (20, 24) ) with the resulting prime ( 3 ). - **Case 3: p and q are both odd** If both ( p ) and ( q ) are odd, and since ( p ) must be prime, it must be ( p 1 ). This leads to ( 8q - p ) and ( 8q - 1 ). To avoid an extra prime factor, we need ( q^2 - 1 8 ), giving ( q 3 ) and ( c 2 ). This results in the solution ( (a, b) (15, 24) ) with the resulting prime ( 2 ). In conclusion, the solutions are ( (a, b) ) in the set ( { (39, 30), (20, 24), (15, 24) } ).

Note: For a more concise presentation, we can summarize the findings as follows:

For ( p 2 ) and ( q 3 ), the solution is ( (a, b) (39, 30) ) with the resulting prime ( 5 ). For ( p 1 ) and ( q 2 ), the solution is ( (a, b) (20, 24) ) with the resulting prime ( 3 ). For ( p 1 ) and ( q 3 ), the solution is ( (a, b) (15, 24) ) with the resulting prime ( 2 ).