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This number : $2305843008139952128$ is perfect as shown here and it was proved that is a perfect number by Euler without using computer , then

My question here is: How Euler showed that number :$2305843008139952128$ is perfect without using computer calculation ?.

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    Ever heard of Mersenne primes? – 2017-02-23
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    but that is not a prime may you meant mersann formula for perfect number which is 2^(n-1)(2^n-1) – 2017-02-23
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    I never said it is. Yes, I mean that formula. Now, $2^{31}-1=2\,147\,483\,647$, and that is not awfully big. You can check it for primality by hand, given enough time. – 2017-02-23
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    by your idea it is easy to check large mersann prime for the exponent is prime if i understand you well – 2017-02-23
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    http://primes.utm.edu/notes/proofs/MerDiv.html – 2017-02-23
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    @Chappers throw that up as an answer – 2017-02-23
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    Great! We have (that is, Euler had) to check the primes below 46341, and only those of them equal to 1 or 63 mod 248. There are only 84 of them. BTW, Lucas-Lehmer test was not known yet. – 2017-02-23
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    Here : https://en.wikipedia.org/wiki/Mersenne_prime it is mentioned that Euler actually proved $2^{31}-1$ to be prime via trial division (probably with the mentioned restriction) and he probably already knew the form of every even perfect number. – 2017-02-23
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    Of course he knew. _Euclid_ knew that, for God's sake! – 2017-02-23
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    What do you meant by God's sake ? – 2017-02-25

2 Answers 2

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Euler was the one to prove that all even perfect numbers are of the form $$2^{n-1} ( 2^n - 1 )$$ This is known as the Euclid-Euler Theorem, and says that we just have to prove that $2^n-1$ is prime. Your number is $$2^{30} ( 2^{31} - 1 )$$ Which Euler got by just plugging integer $n$ into the equation above. Now, we have that $$2^{31}-1=2\,147\,483\,647$$ is small enough to be checked for primality by hand. Euler did this by utilizing modular properties of the Mersenne Primes, as @Chappers and others note in the comments above. In essence Euler lowered the number of cases to check considerably by noting that if $2^{31}-1$ was composite with divisor $p$ then either $p \equiv 1 \pmod {248}$ or $p \equiv 63 \pmod {248}$

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Since the number given is $2^{30}(2^{31}-1)$, all that needs to be done is show that $2^{31}{-}1$ is prime. Euler knew that factors of this number must be of the form $k(2\cdot31)+1$, and also must be $8n\pm1$. This gives $84$ primes to check for division up to $\sqrt(2^{31}-1)$

See Modular restrictions on Mersenne divisors from the Prime Pages.