Question

急求，m平方=m+1，n平方=n+1。求m五次方+n五次方=

Answer 1

m^5
+
n^5

=
m^3
*
m^2
+
n^3
*
n^2

=
m^3
*
(m+1)
+
n^3
*
(n+1)

=
m^4
+
m^3
+
n^4
+
n^3

=
(m+1)^2
+
m(m+1)
+
(n+1)^2
+
n(n+1)

=
m^2
+
2m
+
1
+
m^2
+
m
+
n^2
+
2n
+
1
+
n^2
+
n

=
2m^2
+
3m
+
1
+
2n^2
+
3n
+
1

=
2(m+1)
+
3m
+
1
+
2(n+1)
+
3n
+
1

=
2m
+
2
+
3m
+
1
+
2n
+
2
+
3n
+
1

=
5m
+
3
+
5n
+
3

=
5m
+
5n
+
6

=
5(m+n)
+
6

因为
m^2
=
m
+
1,
n^2
=
n
+
1,
则m,
n为方程
x^2
=
x
+
1的两个根，则有韦达定理，

两根之和
m
+
n
=
-
b/a
=
1,

所以
5(m+n)
+
6
=
5
*
1
+
6
=
11.

所以

m^5
+
n^5

=
5(m+n)
+
6

=
5
*
1
+
6

=
11