Solve for $a,b,c$ given $1/a + 1/b = 1/c$ and a couple other constraints Suppose $a, b, c \in \{4, 5, \ldots, 21\}$ and satisfy $1/a + 1/b = 1/c$. Suppose also that $\gcd(b,c)=1$. Find $a, b, c$.

I tried this problem and I got the answer,
but I can't to understand how $\gcd(b-c,c^2)=\gcd(b,c)$. when gcd(b,c)=1

My attempt:
From question we can see that $b &gt; c$. On
simplifying, we get $$a = c + \frac{c^2}{b-c}.$$
Obviously $b&gt;c$. Multiply both sides by $abc$ to get$$bc+ac=ab$$It follows immediately that$$a=\frac{bc}{b-c}=\frac{bc-c^2+c^2}{b-c}=c+\frac{c^2}{b-c}$$So $b-c|c^2$, but $\gcd(b-c, c^2)=1$ since $\gcd(b, c)=1$. Hence we must have $b-c=1$, which in turn gives $a=c+c^2$.

From $c+c^2=a&lt;22$ one gets $c&lt;5$, so $c=4$ and thus the only solutions is$$(a, b, c)=(c+c^2, c+1, c)$$Where $c=4$.
As you say, $a,b \gt c$.  Presumably you demand $a \neq b$ or there would be many solutions with $a=b=c/2$.  We can demand $a \gt b$ and just try values for $b-c$.  If $b-c=1$, we have $a=c+c^2$ and the only value in range is $c=4, a=20, b=5$.  If $b-c=2,$ we must have $c$ even and $a=c+c^2/2$, which works with $c=4, a=12, b=6$, but $c=6$ makes $a$ too large.  $b-c=3$ gives $a=c+c^2/3$ and gives $c=6, b=9, a=18$.  $b-c=4$ fails because it either gives $c=4, a=8, b=8$ or $c=8, a=24$.  All higher $b-c$ also fail (you need to prove this), so we have three solutions.