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Disprove the following equation

$3987^{12} + 4365^{12} = 4472^{12}$

First, since both the two numbers on the LHS were odd and the RHS was even, I tried dividing by 3 and found

$3 \mid 3987$ and $3 \mid 4365$

so

$3 \mid 3987^{12}$ and $3 \mid 4365^{12}$

and by Divisibility of integer combinations we have

$3 \mid (3987^{12}x + 4365^{12}y)$

For the RHS, $4472$ being even will obviously not be divisible by 3

4472/3 = 1490 with remainder 2

So, $3 \nmid 4472$ and thus $3 \nmid 4472^{12}$

In particular let $x = 1$ and $y = 1$

we will get

$3 \mid (3987^{12} + 4365^{12})$

But we have $3987^{12} + 4365^{12} = 4472^{12}$

so

$3 \mid 4472^{12}$ # a contradiction

thus the original statement is false

Is this a valid proof?

Is it safe to assume this step? $3 \nmid 4472 \Rightarrow 3 \nmid 4472^{12}$

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  • $\begingroup$ Just a fun fact, I found out that this equation is actually correct to the 10th decimal, which is such a near miss to disproving Fermat's Last Theorem. Good job Homer Simpson $\endgroup$ – blablabla Nov 5 '16 at 4:26
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    $\begingroup$ I count only 8 zeros: $(3987^{12} + 4365^{12})^{1/12} = 4472.00000000705\dots$ $\endgroup$ – Tito Piezas III Nov 5 '16 at 4:35
  • $\begingroup$ Oops, my bad... $\endgroup$ – blablabla Nov 5 '16 at 4:36
  • $\begingroup$ @TitoPiezasIII doesn't that technically count as proof that the equation is false? ;D $\endgroup$ – 2012ssohn Nov 5 '16 at 4:38
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    $\begingroup$ You prove is fine but it has 5 times more extraneous information than it needs. Simply state: "3 divides 3987 and 4365 so 3 divides 3987^12+4365^12. But 3 does not divide 4472 so 3 does not divide 4472^12." That'all. And I have 348 characters left. $\endgroup$ – fleablood Nov 5 '16 at 5:13
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The statement '$4472$ being even will obviously not be divisible by $3$' is false (consider $6$). Aside from that, the reasoning is correct. If a prime $p$ does not divide $n$ it divides no power of $n$.

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  • $\begingroup$ I see, if I omit that sentence, everything is good right? Since I have shown that 4472 is not divisble by 3 $\endgroup$ – blablabla Nov 5 '16 at 4:31
  • $\begingroup$ Yes, it is. That said, right now the proof is exceedingly long and more complicated than it needs to be. $\endgroup$ – Fimpellizieri Nov 5 '16 at 4:32
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Is it safe to assume if $n\not |m$ then $n\not |m^k$?

If $n $ has any prime factors that $m $ doesn't, then absolutely!

But if every prime factor of $n $ is a prime factor of $m $ then.... it depends on if the power of $k$ will make the prime factors of $m^k$ higher than the power of $n $.

Example:

Let $n=12=2^2*3$ and $m=18=2*3^2$. $n \not | m $. They have the same prime factors $2,3$ but $n $ has a higher power of $2$ than $m$ does.

But $m^3$ multiplies those powers three fold: $m^3=2^3*3^6$ so now $n|m^3$.

But if $n$ has ANY prime factor, $p $ that $m$ does not, $n\not |m$ because $m$ has no $p $ factor. Raising $m $ to any power isn't going to add any more new prime factors.

Let $4472 =p^aq^b $ be the prime factorisation. We don't actually care what it is (it's something like $4472=8*559$ or something.... I don't care). The only thing we care is that $3$ isn't in it. So $4472^{12}=p^{12a}q^{12b} $ and $3$ didn't slip in. Nothing slipped in. So we know $3\not |4472^{12}.$

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$\gcd(3987,13)=\gcd(4365,13)=1$

$\gcd(4472,13)=13$

$3987^{12} + 4365^{12} \equiv 1 + 1 \equiv 2 \pmod{13}$

$4472^{12} \equiv 0 \pmod{13}$

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