Prove that every even perfect number is a triangular number.even perfect numbers and primesDiscussion on even...
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Prove that every even perfect number is a triangular number.
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Prove that every even perfect number is a triangular number.
even perfect numbers and primesDiscussion on even and odd perfect numbers.Prove that if $2^{p}-1$ is prime then $n=2^{p-1}(2^p-1)$ is a perfect numberWhy does not the perfect number formula imply there are infinitely many perfect numbers?Relationship between Mersenne Primes and Triangular / Perfect NumbersIf n>6 is an even perfect number, Prove that n is congruent to 4 (mod12)Has it been proved that odd perfect numbers cannot be triangular?Even perfect numbers and a relationship with polygonal numbersMultiple of a Triangular Number is another Triangular NumberIf $N = prod_{i=1}^{omega(N)}{{p_i}^{alpha_i}}$ is an odd perfect number, then ${p_i}^{alpha_i} < sqrt{N}$.
$begingroup$
I know a triangular number is given by the formula $frac{n(n+1)}{2}$
I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.
Please help me to prove this.
elementary-number-theory prime-numbers
edited 1 hour ago
Anirban Niloy
666218
asked 3 hours ago
Jake GJake G
442
$endgroup$
add a comment |
$begingroup$
I know a triangular number is given by the formula $frac{n(n+1)}{2}$
I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.
Please help me to prove this.
elementary-number-theory prime-numbers
edited 1 hour ago
Anirban Niloy
666218
asked 3 hours ago
Jake GJake G
442
$endgroup$
add a comment |
$begingroup$
I know a triangular number is given by the formula $frac{n(n+1)}{2}$
I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.
Please help me to prove this.
elementary-number-theory prime-numbers
edited 1 hour ago
Anirban Niloy
666218
asked 3 hours ago
Jake GJake G
442
$endgroup$
I know a triangular number is given by the formula $frac{n(n+1)}{2}$
I also know that an even perfect number is given by $2^text{n-1}(2^text{n}-1)$ if $(2^n-1)$ is prime.
Please help me to prove this.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
edited 1 hour ago
Anirban Niloy
666218
asked 3 hours ago
Jake GJake G
442
edited 1 hour ago
Anirban Niloy
666218
asked 3 hours ago
Jake GJake G
442
edited 1 hour ago
Anirban Niloy
666218
edited 1 hour ago
Anirban Niloy
666218
edited 1 hour ago
Anirban Niloy
666218
666218
asked 3 hours ago
Jake GJake G
442
asked 3 hours ago
Jake GJake G
442
asked 3 hours ago
Jake GJake G
442
442
add a comment |
add a comment |
1 Answer
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$begingroup$
You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
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$begingroup$
You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
You should use different variables in the two expressions. An even perfect number is then $2^{k-1}(2^k-1)$ You need to find an $n$ such that $frac 12n(n-1)=2^{k-1}(2^k-1)$. I find the right side quite suggestive.
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 3 hours ago
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
add a comment |
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