Proving that an expression returns a real non-integer number (Number 2)

Question

Let

$$a=443372888629441 = 17*31*41*43*89*97*167*331$$ $$b=(3+\sqrt{13})/2$$ $$c=(2+\sqrt{8})/2$$ $$d=(1+\sqrt{5})/2$$

How can you prove that the expression

$$\frac{(b^a-1/b^a)-(c^a-1/c^a)-(d^a-1/d^a)}{a}$$

is a real non-integer number?

Answer 1

Let's look at the equations that each of $b, c, d$ are roots of.

$c =1+\sqrt{2} $ is a root of $c^2-2c-1 = 0 $.

$b =(3+\sqrt{13})/2 $ is a root of $b^2-3b-1 = 0 $.

$d$ is a root of $d^2-d-1 =0 $.

Since the constant term of all these equations is $-1$, and each of values are greater than one, their conjugates (in terms of their equation) have magnitude less than $1$ and are, the negative of their reciprocal.

Let's look at $c$, since it is the simplest.

$c = 1+\sqrt{2}$, $1/c =-1+\sqrt{2}$, so, writing $s = \sqrt{2}$, $c^a =(1+s)^a =\sum_{j=0}^a s^j\binom{a}{j} $ and $1/c^a =(s-1)^a =\sum_{j=0}^a s^j (-1)^{a-j}\binom{a}{j} =-\sum_{j=0}^a s^j (-1)^{j}\binom{a}{j} $ so

$\begin{array}\\ c^a-1/c^a &=\sum_{j=0}^a s^j\binom{a}{j} (1+(-1)^j)\\ &=2\sum_{j=0}^{\lfloor a/2 \rfloor} s^{2j}\binom{a}{2j}\\ &=2\sum_{j=0}^{\lfloor a/2 \rfloor} 2^{j}\binom{a}{2j}\\ &=2(1+\sum_{j=1}^{\lfloor a/2 \rfloor} 2^{j}\binom{a}{2j})\\ &=2(1+a\sum_{j=1}^{\lfloor a/2 \rfloor} 2^{j}\frac1{a}\binom{a}{2j})\\ &=2(1+av_b( a))\\ \end{array} $

where $v_b(a) =\sum_{j=1}^{\lfloor a/2 \rfloor} m^{j}\frac1{a}\binom{a}{2j}) $ is an integer under the assumption that $a \big| \binom{a}{2j}$ for $1 \le j \le \lfloor a/2 \rfloor$.

If a similar result holds for $b$ and $d$, the fraction is

$\begin{array}\\ \frac{2(1+av_b( a))-2(1+av_c( a))+2(1+av_d(a))}{a} &=\frac{2(1+a(v_b( a)-v_c( a)+v_d( a))}{a}\\ &=\frac{2}{a}+(v_b( a)-v_c( a)+v_d( a))\\ \end{array} $

which is not an integer.

To make this complete, I would have to work out the sums for $b$ and $d$ and show that they have the same form. I would also have to show that $a \big| \binom{a}{2j}$ for $1 \le j \le \lfloor a/2 \rfloor$.

But it is late and I am tired, so I'll leave it at this.

Answer 2

$b$ and $-1/b$ are roots of $z^2 - 3 z - 1$, so $u_n = b^n + (-1/b)^n$ satisfies the recurrence $u_n - 3 u_{n-1} - u_{n-2}$, with initial conditions $u_0 = 2$, $u_1 = 3$. I find that the periods of this recurrence modulo the primes $17, 31, 41, 43, 89, 97, 167, 331$ that divide $a$ are $16, 64, 28, 42, 180, 196, 336, 664$ respectively. Since $a \equiv 1$ modulo each of these periods, I conclude that $u_a \equiv u_1 = 3$ modulo each of the primes.

Similarly, $c$ and $-1/c$ are roots of $z^2 - 2 z - 1$, and $c^n + (-1/c)^n$ satisfies the recurrence $v_n - 2 v_{n-1} - v_{n-2}$, with initial conditions $v_0 = 2$, $v_1 = 2$. I find that the periods of this recurrence modulo the same primes are $16, 30, 10, 88, 88, 96, 166, 664$ respectively. Again $a \equiv 1$ modulo each of the periods, so $v_a \equiv v_1 = 2$ modulo each of the primes.

A similar story for $d$ and $-1/d$: they are roots of $z^2 - z - 1$, and the recurrence satisfied by $w_n = d^n + (-1/d)^n$ has periods $36, 30, 40, 88, 44, 196, 336, 110$ respectively, and $a \equiv 1$ modulo each of the periods, so $w_a \equiv w_1 = 1$ modulo each of the primes.

Putting it all together,
$u_a - v_a - w_a \equiv 3 - 2 - 1 \equiv 0$ mod each of the primes, and therefore mod $a$. So your expression is an integer.