Suppose-that-the-greatest-common-divisor-of-the-positive-integers-a-b-and-c-is-1-and-ab-a-b-c-Prove-that-a-b-is-a-perfect-square-

Question Number 119657 by bemath last updated on 26/Oct/20

S u p p o s e t h a t t h e g r e a t e s t c o m m o n d i v i s o r o f

t h e p o s i t i v e i n t e g e r s a, b a n d c i s 1 a n d

\frac{a b}{a - b} = c . P r o v e t h a t a - b i s a

p e r f e c t s q u a r e

Commented by som(math1967) last updated on 26/Oct/20

I think a - b = 1 is a perfect square

If a, b both odd then a - b = even

∴ \frac{ab}{a - b} \notin Z [but c \in Z]

if one of a, b even and other

is odd \frac{ab}{a - b} \in Z only a - b = 1

[G . C . D of a, b, c is 1]

so a - b perfect square .

Commented by bemath last updated on 26/Oct/20

y e s

Answered by 1549442205PVT last updated on 26/Oct/20

Supoose a, b, c \in N^{*} . From the hypothesis

we have \frac{a b}{a - b} = c \Leftrightarrow \frac{a (b - a) + a^{2}}{a - b} = c

\Rightarrow - a + \frac{a^{2}}{a - b} = c \Rightarrow (a - b) (a + c) = a^{2} (★)

Suppose \gcd (a - b, a + c) = d . Then

{\begin{cases} a - b = md \\ a + c = nd \end{cases} (*) with \gcd (m, n) = 1 (d, m, n \in N^{*})

(★) \Rightarrow a^{2} = {mnd}^{2} \Rightarrow mn = {(\frac{a}{d})}^{2} = p^{2} (p \in N^{*})

and a^{2} = {(pd)}^{2} \Rightarrow a = pd, mn = p^{2} (1)

Since m, n are coprime, from (1) we infer

there esixts u, v \in N^{*} so that

m = u^{2}, n = v^{2} . Hence, from

(1) and (*) we infer a ⋮ d, b ⋮ d, c ⋮ d

, but by the hypothesis \gcd (a, b, c) = 1

\Rightarrow d = 1, so a - b = md = u^{2} . Thus, a - b is

perfect square (q . e . d)