so for to be an integer must be a multiple of ,
but as this is even must be a multiple of .
But as ranges over the integers (which is the range
that is constrained to) ranges over .
Of the numbers exactly are divisible by 7,
so there are positive integer pairs that satisfy the
given equation.
RonL
Hello, Dragon!
My approach is a variation of CaptainBlack's . . .
How many pairs (x,y) of positive integers satisfy
We have: .
Since is a positive integer, we have:.
There are: multiples-of-seven which are less than 1000
. . and half of them are even.
Therefore, there are 71 solutions.
We can solve this using Bezout's Identity.
Trivially we see that,
Thus,
.
Hence, all integer solutions are given by,
.
We want positive solutions, meaning,
Since, is an integer, we need to find all integers so that,
We see that,
Are all these integers.
In total we have 71 positive solutions.