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
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,
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.