Integer Sum of Powers
If A = B and A n + B n is an integer for all positive integers n , must A and B both be integers?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
4 solutions
Hint: Quadratic formula gets you part of the way. Getting rid of the 2 in the denominator can be slightly tricky.
Log in to reply
Ah, getting rid of 2 in the denominator is tricky, because it cannot be done!
As an example, with a = 2 1 + 5 , b = 2 1 − 5 results in an integer a n + b n . Guess I fooled you (and myself) again.
Log in to reply
We could use use a = 2 + 2 0 , b = 2 − 2 0 and not have the pesky 2 as the denominator.
Log in to reply
@Michael Mendrin – But the phrasing used is "can only". IE It seems like you were trying to classify all potential counterexamples (which I was).
[This is not a complete solution.]
Hint:
First show that
a
,
b
are of the form
2
Z
±
Z
. Why?
Hint:
Counterexamples are pretty easy after that.
Classifying all possible counterexamples
[This part is definitely incomplete. It's just my initial thoughts.]
Claim: The counter examples are of the form 2 p ± q , where any odd-divisor of q is equal to 1 ( m o d 4 ) .
Let
a
+
b
=
x
,
a
2
+
b
2
=
y
, then
a
b
=
2
x
2
−
y
,
∣
a
−
b
∣
=
2
y
−
x
2
and
a
,
b
=
2
x
±
2
y
−
x
2
.
a
n
+
b
n
=
(
a
n
−
1
+
b
n
−
1
)
(
a
+
b
)
−
(
a
b
)
(
a
n
−
2
+
b
n
−
2
)
.
Hence, by induction, if
a
+
b
and
a
b
are integers, then so are
a
n
+
b
n
.
If
a
b
=
2
x
2
−
y
is not an integer, then it is a half integer. In order for
a
n
+
b
n
to always be an integer, it must always be even.
The explicit formula for the Lucas numbers (just like the Fibbonacci numbers except starting with 1,3 instead of 1,1) is L ( n ) = ( 2 1 + 5 ) n + ( 2 1 − 5 ) n
For all A, B satisfying ( A+B = m, AB = n : don't need to be integer )
Let A^n + B^n = S_n
Because A, B are solutions of " X^2 - mX + n = 0 " ,
S n+2 = m S n+1 - n S_n
So, if S 1 and S 2 are integers , all 'S_n's are integer, too.
While I understand what you're going for, unfortunately, it isn't exactly expressed in your solution.
What you currently have is "If A,B are not-integers that satisfsy A+B , AB are integers, then A^n+B^n will always be an integer".
As such, you need to demonstrate that the conditional is true (or at least offer up an explanation of why it is true).
Log in to reply
If A, B are solution of X^2 - mx + n = 0 : ( for example, x^2 - 3x +1 = 0 )
A^n+2 + B^n+2 = m ( A^n+1 + b^n+1) - n ( A^n + B^n )
So, if A+B and A^2 + B^2 are integers, by induction, A^n + B^n is also integer.
( for example, A+B = 3, AB = 1 : A, B : not integer,
but, A^2 + B^2 = 7 : integer,
A^3 + B^3 = 3 ( A^2 + B^2 ) - (A+B) : integer,
A^4 + B^4 = 3 ( A^3 + B^3 ) - (A^2 + B^2 ) : integer,
...
A^n + B^n : integer.
Log in to reply
Right, so what you're missing from the original solution is just saying
( for example, x^2 - 3x +1 = 0 )
That demonstrates that the conditional is true.
Let A = x + y and B = x − y and find out what happens with different n . The odd powers of the radical term cancel out.
Now it does seem restrictive that A , B can only have the form m ± n where m , n are integers