Given a number, $N = 55^5 + 17^5 – 72^5$, then how can we prove that the number is divisible both $3$ and $17$ in a simple way?

Question

I came across this question as a MCQ among which the correct option was that this number is divisible by both $3$ and $17$.

Now one way to show that this number is indeed divisible by both $3$ and $17$ by using the principle of cyclicity of remainders for both the numbers i.e. $3$ and $17$ but that is very time consuming and also cumbersome for such big numbers. So I was wondering if there could be any smart and easy to way to show that this number is divisible by aforementioned numbers.

Thanks in advance !

Answer 1

Consider this as a HINT:

I am not aware of how to solve a degree $5$ expression that you have written, but I hope the below helps and gives you a hint. I have done it for a degree $3$ equation, I suppose this method satisfies for all odd powers, but I couldn't prove it (I'm a junior student myself). Hope this helps:

Note that: $55+17-72=0$

Let's consider $x=55$, $y=17$, $z=-72$

Therefore, we get $x+y+z=0$

$$x^3+y^3+z^3 - 3xyz= (x+y+z)(x^2+y^2+z^2 - xy-yz-zx)$$ $$x^3+y^3+z^3-3xyz = S_1(\sum x^2 -S_2)$$ $$x^3+y^3+z^3 = 3xyz$$

[Because $S_1=x+y+z=0$]

So the expression simplifies to:

$$55^3+17^3-72^3 = 3 \cdot 55 \cdot 17 \cdot (-72)$$

(I have considered $3$ degree, like I said above)

So then you can find factors, visibly $3$ and $17$ are factors.

Answer 2

using the principle of cyclicity of remainders for both the numbers i.e. $3$ and $17$ but that is very time consuming and also cumbersome

No, in fact in general it is likely the quickest and easiest method, viz.

$\quad{\rm suppose}\ \ \, N\, = 55^n + 17^n - 72^n_{\phantom{|^|}}\,$ for $\,{\rm\color{#90f}{odd}}\,\ n\ge 1$
$\quad\!\!\bmod 17\!:\ N\,\equiv\, \color{#c00}{4^n}\, +\ 0^n\, -\ \color{#c00}{4^n}\equiv\, 0_{\phantom{|^|}}$
$\quad\!\!\bmod 3\!:\ \ \ N\,\equiv\ 1^n +\ \color{orange}2^n \ -\ 0^n \equiv\, 0\ $ via $\, \color{orange}2^{n}\equiv (\color{orange}{-1})^n\equiv -1\,$ by $\,n\,\ \rm\color{#90f}{odd}$

Remark $ $ More generally we can exploit the innate modular $\:\!\rm\color{#c00}{symmetry}$ as follows

Theorem $\ $ If $\,\ m\mid \color{#c00}{a\!-\!c},\,\ \overline m \mid a\!-\!d\ $ and $\ m,\overline m,b\,$ are pairwise-coprime then

$\qquad\qquad\quad m\overline m\mid a^n\!+b^n\!-c^n\!-d^n\!\iff {\rm lcm}(o_m(d/b),o_{\overline m}(c/b))\mid n$

where $\,o_n(x) :=$ order of $\,x\pmod{\! n}.\,$ For example

$$17\cdot 19\,\mid 20^n+16^n-3^n-1^n \iff {\rm lcm}(\color{c00}2,\color{0a0}2)=2\mid n\qquad$$

See also this proof that $\ 1897\mid 2903^n - 803^n - 464^n + 261^n$.

Using the above Theorem we can easily handle most all "contest" problems enjoying this symmetry. For many worked examples, follow the links (and their links ...).

Answer 3

Hint:

Use congruences and lil' Fermat:

• Mod $3$, $55\equiv 1$, $17\equiv 2$ and $72\equiv 0$, and $2$ has order $2$.
• Mod $17$, $55$ and $72\equiv 4$.