if x=-2 and y=-1 then p^-q-q^p is
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hope it is helpful
Step-by-step explanation:
Solution 12. From p y + q x
= 1 we have py+qx = xy
and then xy qx py+pq = pq.
Factoring the left side of the last equation gives
(x p)(y q) = pq.
Because x and y are positive integers and p and q are prime we must
have x p = p and y q = q
leading to (x, y) = (2p, 2q)
or
x p = q and y q = p
leading to (x, y) = (p + q, p + q)
or
x p = 1 and y q = pq
leading to (x, y) = (p + 1, pq + q)
or x p = pq and y
q = 1 leading to (x, y) = (pq + p, q + 1).
If p = q, then the first two solutions are the same and we have three
distinct solution sets. If p = q, then the four possibilities result in four
different solution sets.
1
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