Question

Let x and y be positive integers such that 1/x + 1/y =1/7. Find all (x,y).​

Answer 1

Step-by-step explanation:

The equation gives is equivalent to the Diophantine equation

7x+7y=xy.(1)

We can see from equation (1) that 7 divides xy . Therefore, 7 (being a prime number) must divide x or y . As x and y plays symmetric roles in eq. (1), we can assume without loss of generality, that 7 divides x . Then,

x=7q,(2)

for some integer q .

Replacing it in eq. (1) we have

49q+7y=7qy ,

or

7q=y(q−1).(3)

From eq. (3) we see that q divides y(q−1) . Is easy to notice that q and q−1 are coprimes, therefore q must divide y . Hence we have y=qp , for some integer p . Eq. (3) becomes

7q=pq(q−1)

or

7=p(q−1).(4)

As 7 is prime, there are two cases

7 divides p , then p=7r .

7 divides q−1 , then q=7s+1 .

We’ll analyze both cases separately.

First case.

Equation (4) becomes

7=7r(q−1)

or

1=r(q−1).(5)

The Diophantine equation (5) has two different solutions,

rr=1,=−1,qq=2=0.(6)(7)

Solution (7) implies that y=qp=0 , which is impossible, as in the original equation appears 1/y . (We should have said that eq. (1) is equivalent to the original equation provided that x and y are not zero.)

Therefore, the only solution from case 1 is r=1 , q=2 , and then p=7 , hence x=14 , y=14 .

Second case

Equation (4) becomes

7=p⋅7s

or

1=ps.(8)

The possible solutions of eq. (8) are

pp=1,=−1,ss=1=−1(9)(10)

Solution (9) is p=1 , s=1 , then q=8 , and x=56 , y=8 .

Solution (10) is p=−1 , s=−1 , then q=−6 , and x=−42 , y=6 .

It exhausts the possible values for x and y , except for interchanges. There are 3 (or 5 , considering interchanges) solutions:

(x,y)∈{(14,14),(56,8),(8,56),(−42,6),(6,−42)}