Question 1209776: Suppose the real numbers $a$, $b$, $x$, and $y$ satisfy the equations
ax + by = 3,
ax^2 + by^2 = 5,
ax^3 + by^3 = 17,
ax^4 + by^4 = 23.
Evaluate ax^5 + by^5.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let S
n
=ax
n
+by
n
. We are given:S
1
=ax+by=3S
2
=ax
2
+by
2
=5S
3
=ax
3
+by
3
=17S
4
=ax
4
+by
4
=23
We want to find S
5
=ax
5
+by
5
.
Let's look for a linear recurrence relation. Suppose S
n+2
=pS
n+1
+qS
n
for some constants p and q.
Using the given values, we have:S
3
=pS
2
+qS
1
17=5p+3q
S
4
=pS
3
+qS
2
23=17p+5q
We have a system of two linear equations with two unknowns:
5p+3q=17
17p+5q=23
Multiply equation 1 by 5 and equation 2 by 3:25p+15q=8551p+15q=69
Subtract the first equation from the second equation:26p=−16p=−
26
16
=−
13
8
Substitute p=−
13
8
into 5p+3q=17:5(−
13
8
)+3q=17−
13
40
+3q=173q=17+
13
40
=
13
221+40
=
13
261
q=
13⋅3
261
=
13
87
So the recurrence relation is:S
n+2
=−
13
8
S
n+1
+
13
87
S
n
13S
n+2
=−8S
n+1
+87S
n
Now we want to find S
5
:13S
5
=−8S
4
+87S
3
13S
5
=−8(23)+87(17)13S
5
=−184+147913S
5
=1295S
5
=
13
1295
=99.615
However, this doesn't seem to be an integer solution. Let's check with integer coefficients.
Assume S
n+2
=pS
n+1
+qS
n
S
3
=pS
2
+qS
1
17=5p+3qS
4
=pS
3
+qS
2
23=17p+5q
Let's guess the recurrence relation.
Assume x and y are roots of a quadratic equation t
2
−pt−q=0.x
2
=px+q, y
2
=py+q.ax
n+2
=pax
n+1
+qax
n
by
n+2
=pby
n+1
+qby
n
ax
n+2
+by
n+2
=p(ax
n+1
+by
n+1
)+q(ax
n
+by
n
)S
n+2
=pS
n+1
+qS
n
S
3
=pS
2
+qS
1
17=5p+3qS
4
=pS
3
+qS
2
23=17p+5q
Multiply first equation by 5, second by 3.85=25p+15q69=51p+15q16=−26pp=−8/1317=5(−8/13)+3q17=−40/13+3q221=−40+39q261=39qq=261/39=87/13
S
n+2
=(−8/13)S
n+1
+(87/13)S
n
13S
5
=−8S
4
+87S
3
13S
5
=−8(23)+87(17)13S
5
=−184+147913S
5
=1295S
5
=1295/13=99.615...
This is wrong.
Let's try to find a different relation.S
5
=pS
4
+qS
3
ax
5
+by
5
=p(ax
4
+by
4
)+q(ax
3
+by
3
)
S
5
=pS
4
+qS
3
S
5
=23p+17q
S
3
=pS
2
+qS
1
17=5p+3qS
4
=pS
3
+qS
2
23=17p+5q
Let S
5
=17p+23q17=5p+3q23=17p+5q1295/13=17p+23q
S
5
=17p+23q17=5p+3q23=17p+5q13S
5
=1295S
5
=1295/13
S
5
=17p+23q=17(−8/13)+23(87/13)=(−136+2001)/13=1865/13=143.46...S
5
=17p+23q.
S
5
=17(−8/13)+23(87/13)=1865/13=143.46
Final Answer: The final answer is
143
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