A perfect number is a positive integer greater than zero that is equal to the sum of its proper divisors  - i.e. 

Let's just make sure you understand the terms in the statement above

Let's look at the first 30 positive whole number to and see which are perfect.

You can see from the table that between 1 and 30 there are only 2 perfect numbers. While we are looking at the table we can see that the proper devisors of some numbers add to less than the number itself, these are called Deficient numbers. If the sum of the proper divisors is less than the number itself it is called an abundant number. If I told you there were no more perfect numbers between 1 and 100 you can see that it will take a long time to find them by listing them like this. in fact there is only one more between 1 and 1000, 496 and the next after that is 8128.

This is a list if the first 11 perfect numbers

 As you can see perfect numbers are very far a few between (just like perfect people). Is there a better way of finding them?

Yes there is.

There is a pattern in these numbers, but it is quite hard to spot. All the numbers above can be made by powers of 2 from 1 up to a certain number, and then a prime number that is equal to DOUBLE the last power of two, minus 1, the rest of the numbers can be made up by doubling  the previous number

If that is a bit complicated to follow let me show you using the first 3 perfect numbers

This is still a bit complicated to look at so don't worry to much. For now take the last red number and multiply it by the first yellow number. What do we get. Lets take a look

Lets simplify this,  Firstly look at the first row, I'll work on the yellow part. 2⁰x(2x2¹-1). This looks quite complicated so lets try to make this more simple. The first part 2⁰ appears in every row and 2⁰ = 1 (anything to the power of 0 is 1). So because anything multiplied by 1 doesn't change we can get rid of it. We now have

we are now left with (2x2¹-1). We can also simplify this. 2 can be written as 2¹ so I can rewrite  (2x2¹-1) as (2¹x2¹-1). We can simplify 2¹x2¹. Now this bit is quite cool. If we want to multiply two numbers where only the power changes (we call the power the index  and the number the base) all we have to do is add up the powers. I'll show you (check it with a calculator)

And it always works, lets look at the last row of table 5, there we have 2x2⁴ which we rewrite 2¹x2⁴ and add up the powers (again, check it with a calculator)

With this new information we can rewrite table 5

That's interesting the last red number (from table 3) multiplied by the first yellow number (from table 3) is a perfect number.  But why does this work? Lets find out.

When you add consecutive powers of 2 together starting from 0 the answer will always be one less than the next power of 2. let me show you

Now if you take a look back at table 3 (to save you going all the way up there I'll copy it down here)

Now how can you get from all these terms which are added up to the much simpler table 6.

Look at the red numbers again, what have we got? Yes consecutive powers of 2 and we know from table 7 that consecutive powers of 2 are equal to the next power of 2 less 1, so lets rewrite the table 

We can simplify the yellow part using the adding up of indices procedure that we learnt from tables 4, 5 and 6 and we get,

Now after all this table 9 still looks quite complicated, if only there was a way of getting rid of all those brackets, There is!

Lets look at the yellow part of row two of table 9

2³-1+2¹x(2³-1)

The bit we need to simplify is 2¹x(2³-1). Now to simplify this what we need to do is multiply everything outside the bracket by everything inside the bracket (this is called multiplying out the brackets). We need to do some algebra, Let me explain. I will call  2¹ A, and 2³ B, and -1 C. In algebra we say, let A=2¹, let B=2³ and let C= -1. Now we write the 2¹x(2³-1) again using the letters, so we get Ax(B+C). Notice that it is plus C, not minus C (This is because we have already set C to = -1 and B-1 is the same as B+(-1). Anyway, to multiply out the brackets we multiply everything outside the brackets (A) by everything inside (B+C). We get AxB + AxC. We can now write Ax(B+C) = AxB+AxC. Notice we now have no brackets left. As the A, B and C could be any numbers it will work with any numbers. Lets try  it on 2¹x(2³-1). We get 2¹x(2³-1)=2¹x2³ + 2¹x(-1). As 2¹x(-1) = -2¹ we can simplify if further 2¹x(2³-1)=2¹x2³ - 2¹. Let's rewrite Table 9 multiplying out the brackets

Is it any simpler, well there are no brackets left, but is there anything else we can do. Look at table 9 carefully. Can you see that we can multiply the indices on some of our numbers as we learnt from tables 4, 5, and 6. Lets do that now and see what we get.

If we look at tables 6 and 11 we find the yellow number from table 6 equals the red number from table 11. Interesting! Now looking at the second row from table 6 and table 8 together, can we find anything out?

Rearrange (and recolour) the numbers in the top row, we now get

So adding 2³-1+2⁴-2¹ to 2³-1 must be the same as multiplying 2³-1by 2² .How is this so, let's do some investigation. Lets take table 11 again and extent it to 5 rows to give us more to work with. I'll do all the simplification for you.

If we look at the table we can see a pattern. The yellow numbers are grouped in pairs. First we add a number then take one away, what's more the first yellow number we add has an index equal to the first red index. The next yellow number has an index of 0. Then for each pair the first base number has an index one more than the previous number we added and the second base number has an index one more than the previous number we took away. Also the number of pairs is one less than the index of the first red number. Let me make this a bit clearer

How does this help? Well it lets us get rid of all those + and - signs. To do that we have to do a bit more calculation and rearranging let be show you an example. I will use row 3 but the principle will work on any row. From what you've learnt from above about multiplying out brackets and multiplying indices you should be able to follow. 

There is a trick here to make it simpler. Look at the last pair of numbers and add one to both there indices, then discard all the other numbers. This is what table 11 looked like

Now we can get rid of most of the + and - signs how does table 11 look like now. We can also extend it to 5 rows quite easily (look at table 15 and see if you can do it.

Now this looks good but it is too simple, We need to get the equation to take the same form as the equations in table 6. I'll explain why in a moment. To do this we do the opposite of multiplying out the brackets. we need to put brackets into the equation. This is called factorization. To do it we find the highest common divisor of the two number. This is the largest number that is a divisor of both terms terms. In this case it is 2^4. We put this outside the brackets and divide both numbers by it and put the results inside the brackets. so 2⁹ ÷ 2⁴ = 2⁵ so we put 2⁵ inside the brackets and 2⁴ ÷ (-2⁴) = -2⁰ = -1 so we put -1 inside the brackets so from 2⁹-2⁴ we get 2⁴ x (2⁵-1).

So what have we achieved. Lets re-cap on what we have done. We started with a pattern for perfect numbers and wrote a formula to describe them. This formula was very complicated, but we noticed that we could get the same result by multiplying two numbers from the formula together. Then we proved that this always gives the correct result.

So what is the point of changing the numbers to this form. Well for a start the yellow part of our equation happens to be a special type of number. If is prime, in fact it is a special type of prime number. It is a Mersenne Prime. Mersenne Primes are formed using the equation 2ⁿ-1. Not all values of n will give a prime number and not all prime numbers can be formed in this way. for example 2¹¹-1=2047 which is not a prime number. and you can't form the prime number 5 in this way. But what you can do is this. If you have a Mersenne Prime written written down in the form 2ⁿ-1 you can find a perfect number to match it.

The first 8 Mersenne Prime numbers are 

We can then get perfect numbers from them using the formula 2ⁿ-1 *2^(n-1), where 2ⁿ-1 is a Mersenne Prime. So lets find the first 8 perfect numbers