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The problem is to find the number of zeros at the end of the sum $4^{5^6}+6^{5^4}$.

I tried $2^{2 \cdot 5^6}+3^{5^4} \cdot 2^{5^4}= 2^{5^4} \cdot ( 2^{2 \cdot 5^6 -5^4}+ 3^{5^4} )$.

user21820
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piteer
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1 Answers1

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Use the following lemma, which can be proven by induction 'on $b$':

For any odd prime $p$ and $a,b \in \mathbb{N}_{>0}$ such that $p^a \mid\mid x-y$ and $p^b \mid\mid k$:

  $p^{a+b} \mid\mid x^k-y^k$.

For this question we don't need the "strictly divides" in the conclusion, but in general this lemma is very useful.

$4^{5^6}+6^{5^4} = (4^{5^2})^{5^4}+6^{5^4} = (4^{5^2})^{5^4}-(-6)^{5^4}$ and $5 \mid 4^{5^2}-(-6)$ by Fermat's little theorem.

Therefore by the lemma $5^5 \mid 4^{5^6}+6^{5^4}$.

Combining with what you already have gives the number of zeros.

user21820
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  • What's the difference between $||$ and $|$? And how do you read it? – GFauxPas Jan 18 '15 at 14:57
  • $p^k \mid\mid x \Leftrightarrow p^k \mid x \wedge p^{k+1} \nmid x$. – user21820 Jan 18 '15 at 14:57
  • @GFauxPas: Note also that the lemma is slightly different if $p$ is the special even prime $2$. In particular the conclusion would be $p^{a+b+1} \mid\mid x^k-y^k$ if I remember correctly. – user21820 Jan 18 '15 at 14:59
  • @GFauxPas: Read "$\mid$" as "divides" and "$\mid\mid$" as "strictly divides", and note that "strictly divides" only makes sense for (prime) powers. It's technically a shorthand abuse of notation but makes it easy to remember. – user21820 Jan 18 '15 at 15:01