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.

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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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I can give you a proof .

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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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