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It's not by induction, but an easy proof of Fermat's little theorem would be that $x^p \equiv (1+1+1+...+1)^p \equiv (1^p+1^p+...+1^p) \equiv x \pmod p$ with $x$ times number $1$. This is possible because in Pascal's triangle on prime rows the numbers are multiples of p except for the first and last terms which are $1$. This can be easily proven by using binomial formula.

Took some time to realize...
The actual answer is $1921.99999995586722540291132837029507293441170657370868230...$
Unfortunately, most calculators round the answer.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestAre you looking for a simple proof of Fermat's Little Theorem or Fermat's Last Theorem or one of the many other theorems named after Fermat?

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FERMATS LAST THEOREM

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I'm pretty sure Fermat's Last Theorem

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Fermat's Little Theorem can be proved using induction.

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Can u prove it by induction plz show?

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It's not by induction, but an easy proof of Fermat's little theorem would be that $x^p \equiv (1+1+1+...+1)^p \equiv (1^p+1^p+...+1^p) \equiv x \pmod p$ with $x$ times number $1$. This is possible because in Pascal's triangle on prime rows the numbers are multiples of p except for the first and last terms which are $1$. This can be easily proven by using binomial formula.

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Okay get your calculators and try this:

$\sqrt[12]{1782^{12}+1841^{12}}$

$=1922$ right?

So this implies that ${1782^{12}+1841^{12}=1922^{12}}$

Does this disprove Fermat's Last Theorem?

Of course not!

The calculator is wrong.

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BTW: A quick check to see that $1782^{12}+1841^{12} \neq 1922^{12}$ is to note that the left side is odd whereas the right side is even.

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Took some time to realize... The actual answer is $1921.99999995586722540291132837029507293441170657370868230...$ Unfortunately, most calculators round the answer.

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If I remember correctly, that "equation" was from a Homer Simpson episode.

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I can give you a proof .

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