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Mathematics
Problem
Theorem 1
If
(
P
+
Q
)
n
=
I
+
f
(\sqrt{P} + Q)^n = I + f
(
P
+
Q
)
n
=
I
+
f
, where
I
I
I
and
n
n
n
are positive integers,
n
n
n
being odd and
0
≤
f
<
1
0 \le f < 1
0
≤
f
<
1
, then show that
(
I
+
f
)
f
=
k
n
(I + f)f = k^n
(
I
+
f
)
f
=
k
n
, where
P
−
Q
2
=
k
>
0
P - Q^2 = k > 0
P
−
Q
2
=
k
>
0
and
P
−
Q
<
1
\sqrt{P} - Q < 1
P
−
Q
<
1
.
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