@Tushar Gautam
–
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.
–
Michael May
·
3 years, 1 month ago

Log in to reply

I can give you a proof .
–
Ukkash Asharaf
·
1 year, 11 months ago

Log in to reply

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.
–
Ching Z
·
3 years, 1 month ago

Log in to reply

@Ching Z
–
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.
–
Jimmy Kariznov
·
3 years, 1 month ago

Log in to reply

@Ching Z
–
Took some time to realize...
The actual answer is \(1921.99999995586722540291132837029507293441170657370868230...\)
Unfortunately, most calculators round the answer.
–
Vincent Tandya
·
3 years, 1 month ago

Log in to reply

@Ching Z
–
If I remember correctly, that "equation" was from a Homer Simpson episode.
–
Daniel Liu
·
3 years, 1 month ago

## Comments

Sort by:

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? – Jimmy Kariznov · 3 years, 1 month ago

Log in to reply

– Hardik Chandak · 7 months, 3 weeks ago

FERMATS LAST THEOREMLog in to reply

I'm pretty sure Fermat's Last Theorem – Hahn Lheem · 3 years, 1 month ago

Log in to reply

Fermat's Little Theorem can be proved using induction. – Michael Tang · 3 years, 1 month ago

Log in to reply

– Tushar Gautam · 3 years, 1 month ago

Can u prove it by induction plz show?Log in to reply

– Michael May · 3 years, 1 month ago

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.Log in to reply

I can give you a proof . – Ukkash Asharaf · 1 year, 11 months ago

Log in to reply

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. – Ching Z · 3 years, 1 month ago

Log in to reply

– Jimmy Kariznov · 3 years, 1 month ago

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.Log in to reply

– Vincent Tandya · 3 years, 1 month ago

Took some time to realize... The actual answer is \(1921.99999995586722540291132837029507293441170657370868230...\) Unfortunately, most calculators round the answer.Log in to reply

– Daniel Liu · 3 years, 1 month ago

If I remember correctly, that "equation" was from a Homer Simpson episode.Log in to reply