Explain the Fermat's theorem.
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You want Fermat's Last Theorem?
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Fermat's theorem states that if p is a prime number and a is an integer, then for the integer is a multiple of p.
Consider a = 2 and p = 7
Then,
128 - 2 = 7 x 18 is the multiple of 7.
If a is not divisible by p, the theorem is equal to the statement that is a multiple of p i.e., |p| Example: Consider a = 2 and p = 7 then = 64 and 64 − 1 = 63 is a multiple of 7.
Consider a = 2 and p = 7
Then,
128 - 2 = 7 x 18 is the multiple of 7.
If a is not divisible by p, the theorem is equal to the statement that is a multiple of p i.e., |p| Example: Consider a = 2 and p = 7 then = 64 and 64 − 1 = 63 is a multiple of 7.
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FERMAT"s therorem :- if you know about Pythagoras theorem , then you noticed that many triplet can be possible of natural numbers in the form of ,
a² + b² = c²
where a , b and c are natural numbers
but when we increase power of natural number in the same form , we find that
LHS ≠ RHS .
so, theorem is
" if n is natural number greater then 2
then , aⁿ + bⁿ ≠ cⁿ when a , b anc c are also natural numbers."
Let see example :-
take a = 1, b = 2 , c = 3 and n = 3
LHS = aⁿ + bⁿ = (1)³ + (2)³ = 1 + 8 = 9
RHS = cⁿ = (3)³ = 27
LHS ≠ RHS
e.g aⁿ + bⁿ ≠ cⁿ where , n >2 and {a, b, c} € Natural number .
a² + b² = c²
where a , b and c are natural numbers
but when we increase power of natural number in the same form , we find that
LHS ≠ RHS .
so, theorem is
" if n is natural number greater then 2
then , aⁿ + bⁿ ≠ cⁿ when a , b anc c are also natural numbers."
Let see example :-
take a = 1, b = 2 , c = 3 and n = 3
LHS = aⁿ + bⁿ = (1)³ + (2)³ = 1 + 8 = 9
RHS = cⁿ = (3)³ = 27
LHS ≠ RHS
e.g aⁿ + bⁿ ≠ cⁿ where , n >2 and {a, b, c} € Natural number .
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