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How to prove $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$ for all positive integers $a,b,c,d\in \mathbb{N}$? I tried to reduce $d$ to $2$ by induction. But I kind of stuck at the rest of the parameters. Any thoughts?

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1 Answer 1

up vote 4 down vote accepted

Clearly, this is true if at least one of $b,c,d$ is $1$, so we can safely focus on $b,c,d>1$.

$2^{2^{2^2}}-2^{2^2}=2^{16}-2^4=2^4(2^{12}-1)=2^4(2^6-1)(2^6+1)=2^4\cdot7\cdot 9\cdot 13\cdot 5$

We know,for any prime $p$ either $p\mid a$ or $(p,a)=1$

So, if $p\mid a,p\mid(a^e-a)$ for $e\ge 1$ else $(p,a)=1,$

(i)For $p=13,\phi(13)=12$

if $13\mid a,13\mid(a^{b^{c^d}}-a^{b^c})$

if $(13,a)=1$ it is sufficient to prove $12\mid(b^{c^d}-b^c)$

Now, $b^{c^d}-b^c=b^c(b^{c^d-c}-1)=b^c(b^{c(c^{d-1}-1)}-1)$

As $c,d>1$ so, $b^2(b^2-1)\mid b^c(b^{c(c^{d-1}-1)}-1)$

Now, if $2\mid b,2^2\mid b^2$ else $b$ is odd$=2c+1$(say),$b^2-1=(2c+1)^2-1=8\frac{c(c+1)}2$

and $3\mid (b-1)b(b+1)$.

So, $12\mid b^2(b^2-1)\implies 12\mid b^c(b^{c(c^{d-1}-1)}-1)$

(ii) For $p=7,\phi(7)=6$

if $7\mid a, 7\mid(a^{b^{c^d}}-a^{b^c})$

if $(7,a)=1,$ it is sufficient to prove $6\mid(b^{c^d}-b^c)$

(iii) For $p=5,\phi(5)=4$

if $5\mid a, 5\mid(a^{b^{c^d}}-a^{b^c})$

if $(5,a)=1,$ it is sufficient to prove $4\mid(b^{c^d}-b^c)$

$2^4$ and $9=3^2$ need different approach as they are powers$(>1)$ of primes.

(iv)For $2^4,$

If $2\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$, hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $2^4$

If $(2,a)=1,\lambda(16)=\frac{\phi(16)}2=4,$ it is sufficient to prove $4\mid(b^{c^d}-b^c)$ where $\lambda$ is the Carmichael Function.

(v)For $9,$

If $3\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$, hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $3^4,$ hence by $3^2=9$

If $(3,a)=1,\phi(9)=6,$ it is sufficient to prove $6\mid(b^{c^d}-b^c)$

So, in all the cases, $a^{b^{c^d}}-a^{b^c}$ is divisible by $2^4, 7, 9, 13, 5;$ hence by $lcm(2^4, 7, 9, 13, 5)=2^4\cdot7\cdot 9\cdot 13\cdot 5=2^{2^{2^2}}-2^{2^2}$

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