Conjugate radical powers: even sums, integer parts, and recurrence relations

Why adding conjugate powers removes the radical

Let $a,b$ be integers with $b\ge0$, and let $n$ be a positive integer. Then

$$\boxed{(a+\sqrt b)^n+(a-\sqrt b)^n\text{ is an even integer}.}$$

For example, with $a=2,b=3$, the first three sums are $4,14,52$. This cancellation can be proved in two equally rigorous ways.

Proof 1: integer coefficients by induction

We prove simultaneously that integers $p_n,q_n$ exist with

$$(a+\sqrt b)^n=p_n+q_n\sqrt b,\qquad(a-\sqrt b)^n=p_n-q_n\sqrt b.$$

For $n=1$, choose $p_1=a,q_1=1$. If the claim holds at $n=k$, multiplying by the corresponding conjugate base gives

$$p_{k+1}=ap_k+bq_k,\qquad q_{k+1}=p_k+aq_k.$$

Both remain integers, and the two expansions retain opposite radical coefficients. Induction proves the claim. Adding the expansions yields $2p_n$, an even integer.

Proof 2: the binomial theorem

Expand both powers with the binomial theorem. Terms containing an odd power of $\sqrt b$ have opposite signs and cancel. Terms with an even exponent $2j$ occur twice, leaving

$$(a+\sqrt b)^n+(a-\sqrt b)^n=2\sum_{j=0}^{\lfloor n/2\rfloor}\binom n{2j}a^{n-2j}b^j.$$

The sum is an integer, so the result is even. The indexed sum specifies every term explicitly; there is no need to regard this proof as less rigorous than induction.

A small conjugate determines an integer part

Suppose also $0<a-\sqrt b<1$. Then

$$0<(a-\sqrt b)^n<1,$$

so, writing the even sum as $2p_n$,

$$2p_n-1<(a+\sqrt b)^n<2p_n.$$

Consequently

$$\boxed{\lfloor(a+\sqrt b)^n\rfloor=2p_n-1,}$$

which is odd. The strict inequalities are what identify the floor; merely knowing the conjugate sum is integral is not enough.

Problem 1: an odd integer part for every positive exponent

Problem. Prove that the integer part of $(7+\sqrt{38})^n$ is odd for every positive integer $n$.

Because $6<\sqrt{38}<7$, we have $0<7-\sqrt{38}<1$. The conjugate sum

$$(7+\sqrt{38})^n+(7-\sqrt{38})^n=2p_n$$

is an even integer by either proof above. Therefore

$$2p_n-1<(7+\sqrt{38})^n<2p_n,$$

and its floor is the odd integer $2p_n-1$.

Problem 2: a negative conjugate and a last digit

Let $\alpha>\beta$ be the roots of $x^2-4x-1=0$, and define $s_n=\alpha^n+\beta^n$.

1. Find $s_1,s_2,s_3$, and a recurrence expressing $s_n$ through the previous two terms.
2. Find the greatest integer not exceeding $\beta^3$.
3. Find the units digit of the greatest integer not exceeding $\alpha^{2003}$.

Part 1: initial values and recurrence

The roots are $\alpha=2+\sqrt5$, $\beta=2-\sqrt5$. Thus

$$s_1=4,\qquad s_2=18,\qquad s_3=76.$$

Since $\alpha+\beta=4$ and $\alpha\beta=-1$,

$$\boxed{s_n=4s_{n-1}+s_{n-2}\qquad(n\ge3).}$$

The recurrence and integer initial values also give an independent proof that every $s_n$ is an integer.

Part 2: watch the sign of the small conjugate

Here $-1<\beta<0$, so $-1<\beta^3<0$. The required greatest integer is $\boxed{-1}$.

When the conjugate is negative, its power changes sign with parity. The floor argument used for a positive small conjugate must be adjusted accordingly.

Part 3: a recurrence modulo ten

Since $2003$ is odd, $-1<\beta^{2003}<0$. Therefore

$$s_{2003}<\alpha^{2003}<s_{2003}+1,$$

and $\lfloor\alpha^{2003}\rfloor=s_{2003}$.

Modulo ten, the initial values and recurrence produce

$$4,8,6,2,4,8,6,2,\ldots.$$

To justify repetition rather than simply guess it, apply the recurrence to a block $(4,8,6,2)$:

$$4\cdot2+6\equiv4,\quad4\cdot4+2\equiv8,\quad4\cdot8+4\equiv6,\quad4\cdot6+8\equiv2\pmod{10}.$$

Thus each four-term block generates the next. Since $2003\equiv3\pmod4$, the required units digit is $\boxed{6}$.

Problem 3: a product relation and a greatest common divisor

The original article also presents a problem from the University of Tokyo's 2017 science-track paper. Put $p=2+\sqrt5$ and

$$a_n=p^n+\left(-\frac1p\right)^n,\qquad n=1,2,\ldots.$$

1. Find $a_1,a_2$.
2. For $n\ge2$, express $a_1a_n$ using $a_{n+1},a_{n-1}$.
3. Prove that every $a_n$ is a positive integer.
4. Find $\gcd(a_n,a_{n+1})$.

Because $(2+\sqrt5)(2-\sqrt5)=-1$, we have $-1/p=2-\sqrt5$. This is the same conjugate pair as in Problem 2, so $a_1=4$ and $a_2=18$.

Let $q=-1/p$. Since $pq=-1$,

$$\begin{aligned} a_1a_n&=(p+q)(p^n+q^n)\\ &=p^{n+1}+q^{n+1}+pq(p^{n-1}+q^{n-1})\\ &=\boxed{a_{n+1}-a_{n-1}.} \end{aligned}$$

Hence $a_{n+1}=4a_n+a_{n-1}$. The positive integer initial values imply, by induction, that all later terms are positive integers.

Finally, the Euclidean algorithm and the recurrence give

$$\gcd(a_{n+1},a_n)=\gcd(a_{n-1},a_n).$$

Repeatedly reducing the index yields

$$\boxed{\gcd(a_n,a_{n+1})=\gcd(a_1,a_2)=\gcd(4,18)=2.}$$

This shows how the same conjugate-power structure supports floor arguments, congruences, integer recurrences, and divisibility proofs. For related techniques, see rational and irrational parts and Diophantine equations.