Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$

Question

So far I've tried:

1) $26460 = 2^2 * 3^3 * 5 * 7^2$
2) $10152 = 2^3 * 3^3 * 47$
3) $27195 - 10887 = 16308 = 2^2 * 3^3 * 151$ (I know $a^8 - b^8\equiv0\pmod{a - b}$)

Therefore I conclude that $27195^8 - 10887^8 + 10152^8$ is divisible by $2^2 * 3^3$, as well is 1).

But what about $5 * 7^2$ part?

These numbers aren't that big. Can you use a computer to solve the problem? Or even a calculator? — Mastrem, Nov 09 '16 at 15:15
$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \mod 5$. — Santiago, Nov 09 '16 at 15:15
Well I've used Wolfram to check if it's indeed divisible (it is), but the proof requires algebraic manipulation, no more no less. Tnx @Santiago, that helps, onto 7 now. — nullpotent, Nov 09 '16 at 15:19
@iccthedral: notice that $10887 - 10152 = 735 = 7 \cdot 105$. — Santiago, Nov 09 '16 at 15:23

lhf · Answer 1 · 2016-11-09T15:29:52.893

3

Let $a=27195, b=10887, c=10152$ and consider this: $$ \matrix{ m &a \bmod m &b \bmod m &c \bmod m \\ 2^2 &3 &3 &0 \\ 3^3 &6 &6 &0 \\ 5\hphantom{^1} &0 &2 &2 \\ 7^2 &0 &9 &9 \\ } $$ Note how for each $m$, we have $b \equiv a$ or $b \equiv c \bmod m$, and the other is $0 \bmod m$.

Therefore, the terms in $27195^8 - 10887^8 + 10152^8$ reduce to $0 \bmod m$.

edited Nov 09 '16 at 15:29

answered Nov 09 '16 at 15:26

lhf

216,483

1

lhf, what does the matrix represent? – nullpotent Nov 09 '16 at 15:29
Haha, this is too clever to be used as my own proof, professor knows my wits :) Thank you nonetheless, I love the construction. – nullpotent Nov 09 '16 at 15:34
@iccthedral It's a classic exploitation of symmetry - I explain it further in my answer (esp. the linked example). Once you see the innate symmetry it's not "too clever" but somewhat trivial. – Bill Dubuque Nov 09 '16 at 15:42

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

Hint $\ $ Exploit innate symmetry using the following observation

$$\begin{eqnarray}a\equiv c,&&\ b\equiv d\!\!\pmod{m}\\ a\equiv d,&&\ b\equiv c\!\!\pmod{k}\end{eqnarray}\, \Rightarrow\,\ e := a^n+b^n-c^n-d^n\equiv 0\!\! \pmod{{\rm lcm}(m,k))}$$

because $\ \ {\rm mod}\ m\!:\ a\equiv c\,\Rightarrow a^n\equiv c^n,\ \ b\equiv d\,\Rightarrow\ b^n\equiv d^n\,\Rightarrow\,e\equiv 0\, $ by Congruence Power Rule.

Similarly $\, {\rm mod}\ k\!:\,\ e\equiv 0.\ $ Thus $\ m,k\mid e\,\Rightarrow\,{\rm lcm}(m,k)\mid e$.

Remark $\ $ More generally, bringing to the fore the innate symmetry yields:

Assume $ \{a,b\}\equiv \{c,d\}\ {\rm mod}\ m,k\ $ and $\,f(x,y)\,$ is symmetric $\,f(x,y)=f(y,x).$

Then $\,\ \ f(a,b)\equiv f(c,d)\ {\rm mod}\ m,k\,$ so also mod $\,{\rm lcm}(m,k).$

OP has $\, f(x,y) = x^n + y^n.$

Remark $\ $ See here and here for some explicit examples.

Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$

2 Answers2