odd. Hence +1
r
+
1
is even, and one of
p
and
q
must be 2
2
. We may assume that =2
p
=
2
. Then we get 2=+1
2
q
=
r
+
1
and
8+22=2+1=(+1)(−1)+2=2(2−2)+2,
8
+
2
q
2
=
r
2
+
1
=
(
r
+
1
)
(
r
−
1
)
+
2
=
2
q
(
2
q
−
2
)
+
2
,
implying 22−4−6=0
2
q
2
−
4
q
−
6
=
0
or (−3)(+1)=0
(
q
−
3
)
(
q
+
1
)
=
0
. This has =3
q
=
3
as only prime solution, hence =3
q
=
3
and =2−1=5
r
=
2
q
−
1
=
5
.
In conclusion, there are two solutions: (,,)=(2,3,5)
(
p
,
q
,
r
)
=
(
2
,
3
,
5
)
and (,,)=(3,2,5)
(
p
,
q
,
r
)
=
(
3
,
2
,
5
)
.
