How many ordered pairs of positive integers (a,b) are there such that (4a-1)/b and (4b-1)/a are both integers?
To find the number of ordered pairs (a, b) such that both (4a-1)/b and (4b-1)/a are integers, we need to analyze the constraints and conditions on a and b separately.
Let's start with (4a-1)/b. For this expression to be an integer, b must divide (4a-1) evenly. In other words, the remainder of (4a-1) divided by b should be zero (0).
Similarly, for (4b-1)/a to be an integer, a must divide (4b-1) evenly. The remainder of (4b-1) divided by a should be zero (0).
Let's consider the first condition: b divides (4a-1) evenly. Since b divides a certain expression, we can rewrite it as (4a-1) = kb, where k is an integer.
Simplifying, we get 4a = kb + 1, or a = (kb + 1)/4.
Similarly, for the second condition: a divides (4b-1) evenly, we have (4b-1) = la, where l is an integer.
Simplifying, we get 4b = la + 1, or b = (la + 1)/4.
Now, using these equations, we can find the values of a and b that satisfy the conditions.
For a to be an integer, kb + 1 must be divisible by 4. This means kb must leave a remainder of 3 when divided by 4. We can see that the possible remainders of kb when divided by 4 are 0, 1, 2, or 3.
When the remainder is 0: kb is a multiple of 4, and thus a multiple of b. If we subtract 1, we get kb - 1, which is 1 less than a multiple of b. So, when the remainder is 0, a cannot be an integer.
When the remainder is 1: kb + 1 leaves a remainder of 1 when divided by 4. This means kb leaves a remainder of 0 when divided by 4. In this case, kb is a multiple of 4, and a = (kb + 1)/4 is an integer.
When the remainder is 2: kb + 1 leaves a remainder of 2 when divided by 4. This means kb leaves a remainder of 1 when divided by 4. In this case, kb is not a multiple of 4, and a is not an integer.
When the remainder is 3: kb + 1 leaves a remainder of 3 when divided by 4. This means kb leaves a remainder of 2 when divided by 4. In this case, kb is not a multiple of 4, and a is not an integer.
So, the only case where a is an integer is when the remainder of kb divided by 4 is 1.
Applying the same logic to b = (la + 1)/4, we find that the only case where b is an integer is when the remainder of la divided by 4 is 1.
Now, we must determine the number of ordered pairs (a, b) that satisfy both conditions.
To satisfy both conditions, a must have a remainder of 1 when divided by 4, and b must also have a remainder of 1 when divided by 4.
The simplest way to find the number of possible values is to analyze the remainders modulo 4.
When a is divided by 4, the possible remainders are 0, 1, 2, or 3. However, we have already determined that a must have a remainder of 1. Therefore, the only possible remainder for a is 1.
Similarly, when b is divided by 4, the only possible remainder is 1.
Since both a and b must have a remainder of 1 when divided by 4, we can conclude that there is only 1 possible pair of positive integers (a, b) that satisfies the given conditions.
In summary, there is only 1 ordered pair of positive integers (a, b) for which both (4a-1)/b and (4b-1)/a are integers.