MHB Solve for Positive Integer Solutions

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The discussion focuses on finding positive integer pairs (a, b) such that (a² + b²) / (a - b) is an integer that divides 1995. By setting (a² + b²) = k(a - b) and manipulating the equation, it is established that k must be a factor of 1995 that allows 2k² to be expressed as a sum of two squares. The only suitable factor is 5, leading to the solution (a, b) = (3, 1). Further solutions arise by multiplying this basic solution by other factors of 1995, resulting in eight valid pairs. The final solutions are (3, 1), (9, 3), (21, 7), (57, 19), (63, 21), (171, 57), (399, 133), and (1197, 399).
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Find all values of $(a,\,b)$ where they are positive integers for which $\dfrac{a^2+b^2}{a-b}$ is an integer and divides 1995.
 
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[sp]If $\frac{a^2+b^2}{a-b} = k$, then $a^2+b^2 = k(a-b)$. Multiply by $4$ and complete the square, to get $(2a-k)^2 + (2b+k)^2 = 2k^2.$ We want to find factors $k$ of $1995 = 3\cdot 5\cdot 7 \cdot 19$ such that $2k^2$ is a sum of two squares. Now the only way that a number can be expressed as the sum of two distinct squares is if it has factors congruent to $1$ mod $4$. The only such factor in $1995$ is $5$. If we put $k=5$ then $2k^2 = 50 = 1^2 + 7^2$. Putting $2a-5=1$ and $2b+5 = 7$, we get the solution $(a,b) = (3,1).$ The only other solutions will occur through multiplying this basic solution by another factor of $1995.$ Those factors are $3,7,19,21,57,133$ and $399$. Thus there are eight solutions altogether namely $$(a,b) = (3,1),\ (9,3),\ (21,7),\ (57,19),\ (63,21),\ (171,57),\ (399,133),\ (1197,399).$$[/sp]
 
Bravo, Opalg! (Cool)(Clapping)(Sun) And thanks for participating!:)
 
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