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Answer 1 · 2017-10-27T15:39:03

The answer is pairs of numbers of the type $3 n \pm 3$ $3n+-3$ .

$a + b = 6 n$ $a+b=6n$
$a - b = 6$ $a-b=6$

if we sum both equations we obtain:

$2 a = 6 (n - 1) \to a = 3 (n - 1) = 3 n - 3$ $2a=6(n-1)->a=3(n-1)=3n-3$

Replacing a in one of the statements we obtain:

#b=6n-a=6n-(3n-3)=3n+3

So we can have pairs (a, b) equal to:

(-3,3); (0,6); (3;9) etc...