# How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?

Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.

I tried hard but nothing seems to work. 🙁

#### Solutions Collecting From Web of "How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?"

First of all, rewrite the equation as others have, to yield $(2a+1)^n+(2b+1)^n = m\cdot 2^n$, or in other words $(2a+1)^n+(2b+1)^n \equiv 0\pmod {2^n}$. Now, if $n=2k$ is even then for the equation to hold $\bmod 2^{2k}$ it must certainly hold $\bmod 4$; i.e., $\bigl((2a+1)^k\bigr)^2 + \bigl((2b+1)^k\bigr)^2\equiv 0\pmod 4$. But this can’t work, because both of the squares on the left must be congruent to $1 \bmod 4$ and so their sum is congruent to $2$. Therefore, $n$ must be odd, say $n=2k+1$.

Now, the left side can be factored using the classic formula for $\frac{x^n-y^n}{x-y}$ (substitute $x=2a+1, y=-(2b+1)$), yielding
$$(2a+1)^{2k+1}+(2b+1)^{2k+1} = (2a+2b+2)\cdot(x^{2k}+x^{2k-1}y+x^{2k-2}y^2+\cdots+y^{2k})$$
But the factor on the right is odd (it’s the sum of $2k+1$ terms each of which is odd), so for the LHS to be $0\bmod 2^n$, we must have $2a+2b+2\equiv 0 \pmod {2^n}$, and this can only be true for finitely many $n$; it becomes impossible as soon as $2^n\gt 2a+2b+2$.

Hint $\$ Suppose that $\rm\:\left(\dfrac{2a+1}{2}\right)^2 + \left(\dfrac{2b+1}{2}\right)^n = m\in \mathbb Z.\:$ Then scaling by $\rm\:2^n\:$ we deduce

$$\rm 2^{n-2}(2a+1)^2 + (2b+1)^n =\, m\, 2^n$$

which yields a contradiction if $\rm\:n > 2\!:\:$ LHS = even + odd = odd, but RHS is even.

There are much more elegant ways to prove the result, but here’s a quasi-experimental approach that shows how you might attack such a problem. (And it actually proves that there are no positive integers $n$ for which the sum is an integer.)

First note that $\left(a+\frac12\right)^2=a^2+a+\frac14$ is always $\frac14$ more than an integer. Thus, $$\left(a +\frac12\right)^2 + \left(b+\frac{1}{2}\right)^n$$ is an integer if and only if $\left(b+\frac12\right)^n$ is $\frac14$ less than an integer, i.e., if and only if there is an integer $m$ such that $\left(b+\frac12\right)^n=m-\frac14$. If this is the case, then $4\left(b+\frac12\right)^n=4m-1$.

1. Is this possible when $n=1$? No: $4\left(b+\frac12\right)^1=4b+2$ is an integer, but it’s an even integer, and $4m-1$ is odd.

2. What if $n=2$? Still no: $4\left(b+\frac12\right)^2=4b^2+4b+1=4(b^2+b)+1$, which is odd, but it’s one more than a multiple of $4$, and $4m-1$ is one less than a multiple of $4$.

3. For $n=3$ matters are still worse: $4\left(b+\frac12\right)^3=4b^3+6b^2+3b+\frac12$, which isn’t even an integer.

4. For $n=4$ we have $4\left(b+\frac12\right)^4=4b^4+8b^3+6b^2+2b+\frac14$, which is worse yet.

Now look at the constant terms in (1)-(4): $2,1,\frac12$, and $\frac14$. They should suggest the conjecture that the constant term in $4\left(b+\frac12\right)^n$ is $\dfrac1{2^{n-2}}$. If you can prove this, you’re practically done. Equivalently, try to prove the

Conjecture: The constant term in $\left(b+\frac12\right)^n$ is $\dfrac1{2^n}$.

If you already know the binomial theorem, you can get this immediately from it. Otherwise, you can prove the conjecture by induction on $n$.

Note that
$$\left(a+\frac{1}{2}\right)^n + \left(b+\frac{1}{2}\right)^n=\frac{1}{2^n}\left(c^n+d^n\right),$$
where $c=2a+1$ and $d=2b+1$.

Let $2^e$ be the highest power of $2$ that divides $c+d$.

If $n$ is even, the highest power of $2$ that divides $c^n+d^n$ is $2$.

For odd $n$, note that $u^n+v^n=(u+v)(u^{n-1}+u^{n-2}v+\cdots +v^{n-1})$. The second term in this product is odd. It follows that the highest power of $2$ that divides $c^n+d^n$ is $2^e$.

Thus if $n\gt e$, then $\left(a+\frac{1}{2}\right)^n + \left(b+\frac{1}{2}\right)^n$ is not an integer.

Notice that
$$\left(a+\frac12\right)^n+\left(b+\frac12\right)^n=\left((2a+1)^n+(2b+1)^n\right)/2^n$$
and use the binomial theorem to expand.

The problem is equivalent to the following:

Let $q\in\mathbb{Q}^{+}$ with $\nu_2(q)=0$. Prove that $\nu_2(q^n+1)\geq n$ for only finitely many $n$.

Proof. If $2|n$ then $\nu_2(q^n+1)\leq 1$. Now choose odd $n$ large enough so that $||q+1||<2^n$ where $||\frac{a}{b}||=|a|+|b|$.
Then $v_2(q^n+1)=v_2(q+1)+v_2\left(\sum_0^{n-1} (-q)^i\right)=v_2(q+1)<n$.

Hence only finitely many solutions exist.

Explanation of my reformulation:
By rearrangement, the initial equation is $(2a+1)^n+(2b+1)^n\in 2^n \mathbb{Z}$.
The pair $(2a+1,2b+1)$ is equivalent to specifying a $q\in\mathbb{Q}^{+}$ with $\nu_2(q)=0$, because we can make $(2a+1,2b+1)\to\frac{2a+1}{2b+1}$ and likewise map $q=\frac{m}{n}\to(m,n)$ with $m,n$ relatively prime.
Then the condition is just saying $\nu_2(q^n+1)\geq n$.