Proving that an expression returns a real non-integer number (Number 2) 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?
 A: 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.
A: $b$ 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.
