Solving the Equation a3 - b3 - ab 25 for Positive Integer Solutions

In this article, we will explore how to solve the equation a3 - b3 - ab 25 for positive integer solutions. We will break down the process into several steps, detailing the algebraic maneuvers involved and providing tests with small integer values to find the correct solutions.

Step-by-Step Solution

To solve the equation, we begin by rewriting it in a more manageable form:

a3 - b3 ab 25

We can then factor the left-hand side using the difference of cubes formula:

(a - b)(a2 ab b2) ab 25

Next, let's denote a - b k. So, a b k. Substituting this into the equation gives:

(b k)kb2 (b k)kb (b k)2 (b k)b2 (b k)b 25

Expanding the left-hand side, we get:

k3b2 3bk2 k2 3kb2 3bk b2 b3 2bk2 k2b 25

After simplifying and combining like terms, the equation becomes:

k3b2 3bk2 k2 3kb2 3bk b2 b3 2bk2 k2b 25

This simplifies further to:

k3b2 3bk2 k2 3kb2 3bk b2 - b3 - 2bk2 - k2b 25

Further simplification yields:

(k - 1)2b (k - 1)3b2 (k - 1)b3 25

Testing for Feasible Values of a and b

Since a and b are positive integers, we can test small values for b and find corresponding a. Here are the tests for the values of b:

Testing b 1:

a3 - 1 - a 25 implies a3 - a - 26 0

Testing a 3:

33 - 3 - 26 27 - 3 - 26 -2, not a solution

Testing a 4:

43 - 4 - 26 64 - 4 - 26 34, not a solution

Testing a 5:

53 - 5 - 26 125 - 5 - 26 94, not a solution

Testing a 6:

63 - 6 - 26 216 - 6 - 26 184, not a solution

Testing b 2:

a3 - 8 - 2a 25 implies a3 - 2a - 33 0

Testing a 4:

43 - 2 * 4 - 33 64 - 8 - 33 23, not a solution

Testing a 5:

53 - 2 * 5 - 33 125 - 10 - 33 82, not a solution

Testing a 6:

63 - 2 * 6 - 33 216 - 12 - 33 171, not a solution

Testing b 3:

a3 - 27 - 3a 25 implies a3 - 3a - 52 0

Testing a 4:

43 - 3 * 4 - 52 64 - 12 - 52 0, a 4, b 3 is a solution

Testing b 4:

a3 - 64 - 4a 25 implies a3 - 4a - 89 0

Testing a 5:

53 - 4 * 5 - 89 125 - 20 - 89 16, not a solution

Testing a 6:

63 - 4 * 6 - 89 216 - 24 - 89 103, not a solution

Testing b 5:

a3 - 125 - 5a 25 implies a3 - 5a - 150 0

Testing a 6:

63 - 5 * 6 - 150 216 - 30 - 150 36, not a solution

Testing a 7:

73 - 5 * 7 - 150 343 - 35 - 150 158, not a solution

Testing b 6:

a 7 gives 73 - 63 - 42 25, solution found

Conclusion

The found pairs are (4, 3) and (7, 6). Now let's calculate a2 b3 for each:

For (4, 3):

42 33 16 27 43

For (7, 6):

72 63 49 216 265

Thus, the largest possible value of a2 b3 is:

boxed{265}