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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?Multiple of a Triangular Number is another Triangular NumberIf $N = prod_{i=1}^{omega(N)}{{p_i}^{alpha_i}}$ is an odd perfect number, then ${p_i}^{alpha_i} < sqrt{N}$.Prove that if $ab$ is a perfect square and $GCD(a,b)=1$, then $a$ and $b$ are perfect squares

4

$begingroup$

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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.

Please help me to prove this.

elementary-number-theory prime-numbers

share|cite|improve this question

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

$endgroup$

add a comment | 

4

$begingroup$

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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.

Please help me to prove this.

elementary-number-theory prime-numbers

share|cite|improve this question

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

$endgroup$

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 $2^{n-1}(2^{n}-1)$ if $(2^n-1)$ is prime.

Please help me to prove this.

elementary-number-theory prime-numbers

share|cite|improve this question

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

$endgroup$

share|cite|improve this question

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

share|cite|improve this question

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

edited 14 hours ago

Vinyl_coat_jawa

3,16711233

asked 18 hours ago

Jake GJake G

613

asked 18 hours ago

Jake GJake G

613

2 Answers
2

active

oldest

votes

6

$begingroup$

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. Therefore $n=2^{k}-1$

share|cite|improve this answer

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  n should be 2^(k)-1
  $endgroup$
  – Jake G
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
  $endgroup$
  – Vinyl_coat_jawa
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
  $endgroup$
  – David Mulder
  12 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You could think as follows:

Every triangular number you can represent as follows:

$$
bullet\
bullet bullet\
bullet bullet bullet \
...
$$

that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.

Multiplying this by $2$ gives a rectangle

$$
bullet circ circ circ\
bullet bullet circ circ\
bullet bullet bullet circ\
$$

with one side exactly one unit longer than the other.

So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.

Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed

$$
frac{n(n+1)}{2}=2^{k-1}(2^k-1),
$$

so
$$
text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

$endgroup$

add a comment | 

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6

$begingroup$

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. Therefore $n=2^{k}-1$

share|cite|improve this answer

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  n should be 2^(k)-1
  $endgroup$
  – Jake G
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
  $endgroup$
  – Vinyl_coat_jawa
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
  $endgroup$
  – David Mulder
  12 hours ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

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. Therefore $n=2^{k}-1$

share|cite|improve this answer

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  n should be 2^(k)-1
  $endgroup$
  – Jake G
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JakeG it is not nice editing the answer you got for one of your questions! Add a comment or something (I saw you did) but then why did you edited your comment into the answer?
  $endgroup$
  – Vinyl_coat_jawa
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vinyl_coat_jawa There is nothing more or less nice about editing an answer to a question you did or did not ask. If you can improve an answer without breaking the author's intent then editing the answer is a good idea. If anything it's the comment that isn't necessary, but considering Jake G doesn't have edit privileges yet and had to go through the review queue doing both made sense.
  $endgroup$
  – David Mulder
  12 hours ago
  
  
  

  
  
  
  

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. Therefore $n=2^{k}-1$

share|cite|improve this answer

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

$endgroup$

share|cite|improve this answer

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

edited 14 hours ago

Jake G

613

edited 14 hours ago

Jake G

613

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

answered 18 hours ago

Ross MillikanRoss Millikan

298k23198371

1

$begingroup$

You could think as follows:

Every triangular number you can represent as follows:

$$
bullet\
bullet bullet\
bullet bullet bullet \
...
$$

that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.

Multiplying this by $2$ gives a rectangle

$$
bullet circ circ circ\
bullet bullet circ circ\
bullet bullet bullet circ\
$$

with one side exactly one unit longer than the other.

So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.

Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed

$$
frac{n(n+1)}{2}=2^{k-1}(2^k-1),
$$

so
$$
text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

$endgroup$

add a comment | 

1

$begingroup$

You could think as follows:

Every triangular number you can represent as follows:

$$
bullet\
bullet bullet\
bullet bullet bullet \
...
$$

that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.

Multiplying this by $2$ gives a rectangle

$$
bullet circ circ circ\
bullet bullet circ circ\
bullet bullet bullet circ\
$$

with one side exactly one unit longer than the other.

So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.

Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed

$$
frac{n(n+1)}{2}=2^{k-1}(2^k-1),
$$

so
$$
text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

$endgroup$

add a comment | 

$begingroup$

You could think as follows:

Every triangular number you can represent as follows:

$$
bullet\
bullet bullet\
bullet bullet bullet \
...
$$

that is $1+2+3+ldots +n=frac{n(n+1)}{2}$.

Multiplying this by $2$ gives a rectangle

$$
bullet circ circ circ\
bullet bullet circ circ\
bullet bullet bullet circ\
$$

with one side exactly one unit longer than the other.

So if you build such a rectangle and you choose one side of the rectangle to be $2^k$ long, the other side will be $(2^k-1)$ long (where you picked $k$ so that $(2^k-1)$ is prime as required.). It will have "area" (i.e. nunber of dots in it) $2^k(2^k-1)$ and the half of this is always a triangular number.

Since you can do this for any $k$ fullfilling the condition that $2^k-1$ is prime you showed

$$
frac{n(n+1)}{2}=2^{k-1}(2^k-1),
$$

so
$$
text{$a$ is perfect}Rightarrow text{$a$ is triangular}.
$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233

answered 13 hours ago

Vinyl_coat_jawaVinyl_coat_jawa

3,16711233