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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 $b > c$. On simplifying, we get $$a = c + \frac{c^2}{b-c}.$$

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5 Answers

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Obviously $b>c$. 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 $c+c^2=a<22$ one gets $c<5$, so $c=4$ and thus the only solutions is$$(a, b, c)=(c+c^2, c+1, c)$$Where $c=4$.

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Then equation:

$$XY+XZ+YZ=N$$

If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written.

$$X=p$$

$$Y=s-p$$

$$Z=k-p$$

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$1/a + 1/b = 1/c$ so

$1/a =1/c - 1/b$ so

$a = 1/[1/c-1/b] = bc/(b-c)$

As this is a whole number $b-c$ divides $bc.$ So any factor of $b-c$ must be a factor of either $b$ or $c.$ If it's a factor of $b$ and also $b-c$ it is a factor of $c$ as well. If is a factor of $c$ and also $b-c$ then it is also a factor of $b.$ Either way, any factor of $b-c$ is a factor of both $b$ and $c.$

But $\gcd(b,c)=1$ so $1$ is the only factor of $b-c.$ So $b-c = \pm 1$ and as $a,b,c$ are positive $b-c=1.$

So $a=bc$ so $a\geq 4\times 5=20.$ $ a\leq 21$ and $21$ is clearly not an option. $(21=3x7 != bc$ where $b-c=1)$

So $a=20, b,c=4,5.$ $ b-c=1 $ so $b=c+1$ so $b=5$ and $c=4.$

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Let $p$ prime $p|(b-c)$ and $p|c^2$. From $p|c^2$ we get $p|c$ then $p|b$ because $b=(b-c) + c$.

Now let $p$ prime $p|b, p|c$. It follows that $p|(b-c), p|c^2$.

The conclusion is $\gcd(b,c)$ and $\gcd(b-c,c^2)$ have the same prime divisors. From here we deduce if one of them is $1$ the other one is also $1$.

Back to your problem. Because $\gcd(b,c)=1$, we have $\gcd(b-c,c^2)=1$

But $$a = c + \frac{c^2}{b-c} \tag1$$ therefore $(b-c)|c^2$. It follows that $b-c=1$ (otherwise, if $b-c \ge2$ then there is a prime $p | b-c$ and because $(b-c)|c^2$ we get $\gcd(b-c,c^2) \ne 1$ false).

Now we'll find all the coprime pairs ($c,b=c+1$) where $c \in \{4, 5, \ldots, 20\}$ satisfying (1).

We get $ c=4, b=5, a=20 $ as unique solution. There is no reason to check for $c \ge 5$ because $c^2 \ge 25$ implies $a \gt 25$, which is false.

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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.

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