Find all integers x for which there exists an integer y such that
1/x + 1/y = 1/7.
I'm not exactly sure how to approach this problem. Please answer as soon as possible.
this might be a good place to start.
google is your friend.
https://web2.0calc.com/questions/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
To solve this problem, we can start by simplifying the equation 1/x + 1/y = 1/7.
To combine the two fractions on the left-hand side, we need a common denominator. In this case, the common denominator is xy.
So, let's multiply both sides of the equation by xy:
xy * (1/x) + xy * (1/y) = xy * (1/7)
After canceling out terms, we get:
y + x = xy/7
Rearranging the equation, we have:
xy - 7y - 7x = 0
Now let's treat this equation as a quadratic equation in terms of y:
y^2 - 7x*y - 7x = 0
Applying the quadratic formula, we have:
y = (7x ± √(49x^2 + 28x))/(2)
Since y is an integer, the discriminant (49x^2 + 28x) must be a perfect square. Let's form an equation from the discriminant:
49x^2 + 28x = d^2
where d is a positive integer.
Now we have a Diophantine equation. To solve this equation, find integer solutions for x such that 49x^2 + 28x is a perfect square.
One approach to solve this equation is to consider the factors of 49 (prime factorization of 49 is 7 * 7).
Let's examine two cases:
Case 1: When 7 divides x.
Let's assume x = 7k, where k is an integer.
Substituting into the equation:
49(7k)^2 + 28(7k) = d^2
(7^2)(7k^2 + 4k) = d^2
This shows that the expression inside the brackets, 7k^2 + 4k, must be a perfect square. Let's call it m^2:
7k^2 + 4k = m^2
We now have another quadratic Diophantine equation. One way to solve it is to complete the square. By doing so, we get:
(2k + 1)^2 - m^2 = 1
This equation is a special case of Pell's equation, which has infinitely many solutions. One such solution is when (2k + 1) = 3 and m = 2:
(2k + 1)^2 - m^2 = 1
(2(1) + 1)^2 - (2)^2 = 1
(3)^2 - (2)^2 = 1
9 - 4 = 1
Therefore, x = 7k = 7(1) = 7 and y = (7x ± √(49x^2 + 28x))/(2) = (7(7) ± √(49(7)^2 + 28(7)))/(2) = 98/2 or -56/2. The integers are x = 7 and y = 49 or -28.
Case 2: When 7 does not divide x.
In this case, let's assume x = 7k + r, where r is the remainder when x is divided by 7.
Substituting into the equation:
49(7k + r)^2 + 28(7k + r) = d^2
(7^2)((7k + r)^2 + 4(7k + r)) = d^2
(7^2)(49k^2 + 14kr + r^2 + 28k + 4r) = d^2
(7k^2 + 14kr + r^2 + 4k + r) = (d^2)/49
Since d^2/49 is a perfect square, we can let (d/7)^2 = n^2:
7k^2 + 14kr + r^2 + 4k + r = n^2
Similar to Case 1, we complete the square:
(k + r/2)^2 - (3r^2 + 4k + r) = 0
Again, we have a Diophantine equation. One possible solution is when k = 1 and r = -1:
(1 + (-1)/2)^2 - (3(-1)^2 + 4(1) + (-1)) = 0
(3/2)^2 - (3 - 4 - 1) = 0
(3/2)^2 - (3 - 5) = 0
(3/2)^2 - (-2) = 0
9/4 + 8/4 = 17/4 = 0
Therefore, x = 7k + r = 7(1) + (-1) = 6 and y = (7x ± √(49x^2 + 28x))/(2) = (7(6) ± √(49(6)^2 + 28(6)))/(2) = 42/2 or 60/2. The integers are x = 6 and y = 21 or 30.
So, the possible integer solutions for x and y are x = 7, y = 49 or -28, and x = 6, y = 21 or 30.