PLEASE SOLVE THIS QUESTION FAST Let a and b be positive integers such that ab + 1 divides a2 + b2. Show that {\displaystyle {\frac {a^{2}+b^{2}}{ab+1}}} is the square of an integer.
Answers
Answer:
Let
a
and
b
be positive integers such that
a
b
+
1
divides
a
2
+
b
2
. Show that
a
2
+
b
2
a
b
+
1
is the square of an integer.
Assuming that
a
2
+
b
2
a
b
+
1
=
n
2
and solving for
a
we get
a
=
1
2
(
b
n
2
±
√
4
n
2
+
b
2
(
n
4
−
4
)
)
and making
n
=
2
m
and substituting
a
=
2
b
m
2
−
√
4
m
2
+
b
2
(
4
m
4
−
1
)
now we need
4
m
2
+
b
2
(
4
m
4
−
1
)
≥
0
or
b
2
≤
4
m
2
1
−
4
m
4
which is unfeasible so we consider now instead
4
m
2
−
b
2
=
0
and then
a
=
2
b
m
2
±
2
b
m
2
or for feasibility
a
=
4
b
m
2
=
4
(
2
m
)
m
2
=
8
m
3
and finally the parameterization
{
a
=
8
m
3
b
=
2
m
for
m
=
1
,
2
,
3
,
⋯
gives us
a
2
+
b
2
a
b
+
1
=
4
m
2
which is a perfect square.
NOTE: A question remains. Covers this parameterization all the possibilities?
Answer:
Explanation:
Let
a
and
b
be positive integers such that
a
b
+
1
divides
a
2
+
b
2
. Show that
a
2
+
b
2
a
b
+
1
is the square of an integer.
Assuming that
a
2
+
b
2
a
b
+
1
=
n
2
and solving for
a
we get
a
=
1
2
(
b
n
2
±
√
4
n
2
+
b
2
(
n
4
−
4
)
)
and making
n
=
2
m
and substituting
a
=
2
b
m
2
−
√
4
m
2
+
b
2
(
4
m
4
−
1
)
now we need
4
m
2
+
b
2
(
4
m
4
−
1
)
≥
0
or
b
2
≤
4
m
2
1
−
4
m
4
which is unfeasible so we consider now instead
4
m
2
−
b
2
=
0
and then
a
=
2
b
m
2
±
2
b
m
2
or for feasibility
a
=
4
b
m
2
=
4
(
2
m
)
m
2
=
8
m
3
and finally the parameterization
{
a
=
8
m
3
b
=
2
m
for
m
=
1
,
2
,
3
,
⋯
gives us
a
2
+
b
2
a
b
+
1
=
4
m
2
which is a perfect square.
NOTE: A question remains. Covers this parameterization