Find all integers x for which there exists an integer y such that 1/x + 1/y = 1/7

(In other words, find all ordered pairs of integers (x,y) that satisfy this equation, then enter just the x's from these pairs.)

Guest Aug 26, 2019

#1**+2 **

1/x + 1/y = 1/7

Since both x and y will be greater than 7, let z =7 and we can write this

1/ ( z + a) + 1 / (z + b) = 1 / z get a common denominator on the right

[ z + b + z + a ] 1

_____________ = ____ simplify

(z + a) (z + b) z

[2z + a + b] 1

_______________ = _____ cross-multiply

z^2 + az + bz+ ab z

2z^2 + az + bz = z^2 + az + bz + ab

z^2 = ab

7^2 = ab

49 = ab

We have these pairs of integers ab that multiply to 49

(a,b) = (-7, -7) So z + a = x = 0 and z + b = y = 0 we cannot divide by 0 so reject this pair

(a, b) = (7, 7) so z + a = x = 14 qnd z + b = y =14 so x = 14 and b = 14

(a, b) = (1, 49) so z + a = x = 8 and z + b = = y = 56 so x = 8 and b = 56

So

(x, y) = ( 14, 14) and ( 8, 56) or (56, 8)

CPhill Aug 26, 2019