Pairs of Positive Integers

How many pairs of positive integers \left (a,b\right ), with a and b both less than 100, satisfy the equation a\sqrt{2a+b}=b\sqrt{b-a}?
Source: NCTM Mathematics Teacher

\sqrt{b-a} implies that b-a\geqslant 0 or b\geqslant a.
Suppose b=a
a\sqrt{3a}=0  Not possible because a is a positive integer.
We conclude that b>a. Since b is a positive integer, there exists some positive integer x such that b=a+x.
a\sqrt{2a+\left (a+x\right )}=\left (a+x\right )\sqrt{\left (a+x\right )-a}
a\sqrt{3a+x}=\left (a+x\right )\sqrt{x}
Squaring both sides of the equation
a^2\left (3a+x\right )=x\left (a+x\right )^2
Divide both sides by 3a+x\neq 0
a^2=\displaystyle\frac{x\left (a+x\right )^2}{3a+x}
a is an integer implies that a^2 is an integer. This in turn requires that 3a+x divides x\left (a+x\right )^2 which only occurs when x=a.
\displaystyle\frac{a\left (a+a\right )^2}{3a+a}=\frac{a\left (4a^2\right )}{4a}
The answers are 49 pairs \left (a,2a\right )
\left (1,2\right ),\left (2,4\right ),\left (3,6\right ),\dots,\left (49,98\right )

Answer: 39

Alternative solution
Squaring both sides
a^2\left (2a+b\right )=b^2\left (b-a\right )
Rearrange the terms
Factor both sides
a\left (a^2+ab+b^2\right )=\left (b-a\right )\left (a^2+ab+b^2\right )
Divide both sides by a^2+ab+b^2\neq 0

About mvtrinh

Retired high school math teacher.
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