The only perfect number of the form x^n + y^n
What number is supposed to be "perfect"? x, y or n?
There are an infinite number of numbers that can be written as x^n + y^n, if one is free to pick x, y and n.
Define perfect number if you wish the question answered. It must not be the Pythagorean definition of a perfect number, since there are more than one of them.
I assume that you are seeking values of x and y that make x^n + y^n = N, a perfect number, 2^(p-1)(2^p - 1), where p is a Mersenne prime number.
A perfect number is an integer that is equal to the sum of its positive divisors (not including itself). Therefore, 6 is a perfect number, since its positive divisors are 1, 2, and 3 and 1+2+3 = 6. 28, 496, and 8128 are also perfect numbers. At present, there are over 30 known perfect numbers, all even. All even perfect numbers are of the form 2^(p-1)(2^p - 1), where p is any positive integer, exceeding unity, that makes (2^p -1) is prime. The primes of the form (2^p - 1), where p is a prime, are called Mersenne primes after the French mathematician who, in 1644, announced a list of new perfect numbers. The known values of p that yield Mersenne primes and corresponding perfect numbers are: 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1,279, 2,203, 2,281, 3,217, 4,253, 4,423, 9,689, 9,941, 11,213, and 19,937. Though there are undoubtedly many more beyond p = 19,937, their size grows rapidly as p increases. The 15th Mersenne prime, (2^1,279 - 1) has 386 digits. It is not known if there are any odd perfect numbers; none has been found, but it has not been proved that one cannot exist. The first eight perfect numbers are 6, 28, 496, 8128, 33,550,336, and 8,589,869,056, 137,438,691,328, and 2,305,843,008,139,952,128, having p's of 2, 3, 5, 7, 13, 17, 19, and 31.
I suggest you search through Diophantus' Arithmetica for a method of solving equations of the form x^n + y^n = N. There might be a Pythagorean Triple hypotenuse that is a perfect number, but I doubt it.
Solutions of x^2 + y^2 = N.
Equations of the form ax + by = c, x^2 - y^2 = N, x^2 + y^2 = N, and x^2 - Dy^2 = z^2 appear frequently in the broad field of recreational mathematics. The first three are usually referred to as Diophantine equations and the last, Pell equations. Varying methods exist for their solution. The solutions of ax + by = c and x^2 - y^2 = N appear elsewhere in the Knowledge Database. The following addresses the solution of x^2 + y^2 = N. The equation is often considered applicable to finding the sides of a Pythagorean Triangle given only the integer magnitude of the hypotenuse.
Finding solutions to x^2 + y^2 = N is comparible to sqtrN = z being the hypotenuse of a Pythagorean triangle and being asked to find the two legs of the triangle. We are all familiar with the expressions that define a Pythagorean triple, x = k(m^2 - n^2), y = k(2mn), and z = k(m^2 + n^2) where m and n are two relatively prime positive integers, m greater than n, one odd, one even, and k is any positive integer. When k = 1, the triple is a primitive triple. A primitive triple is one where the three sides have no factor in common.
As with the solutions to x^2 - y^2 = N, the first thing that is often helpful is to recognize whether N is prime, composite, a square, or a higher power. Being odd or even is sometimes important. All in all, we have the possibilities of N being a prime, an odd square, an even square, an odd non square, an even non square, or either an odd or even higher power. Lets examine all of these possibilities.
It is often advantageous to determine whether N is a prime, a square, or non square first, though not always. Toward that end, the following general rules will, in many instances, help in making this determination. Keep in mind that all primes are of the form 4n + 1 or 4n - 1.
1--Does N end in 1, 3, 7 or 9?
If no, N is not a prime
...If yes, N might be a prime.
2--Is N divisible by any other primes less than sqrtN?
...If no, N is a prime.
...If yes, N is a composite.
3--Is (a^N - a) evenly divisible by N?
...If yes, N is a prime.
...If no, N is a composite.
4--Is (N - 1)! + 1 evenly divisible by N?
...If Yes, N is a prime.
...If no, N is a composite.
5--Is [a^(N-1) - 1], evenly divisible by N?
...If yes, N is possibly a prime.
...If no, N is a composite.
6--If F(n) = [cos^2(Pi)(N-1)! + 1]/N = 1, N is a prime.
...If F(n) = [cos^2(Pi)(N-1)! + 1]/N = 0, N is composite.
7--Does the number end in 0, 1, 4, 5, 6 or 9?
...If yes, N might be a square.
...If no, N is not a square.
8--Do the digits of N add up to 1, 4, 7 or 9?
...If yes, N is a square.
...If no, N is not a square.
9--Is N evenly divisible by 4 or leave a remainder of 1 when divided by 4?
...If yes, N is a square.
...If no, N is not a square.
10--Is (N - 1) evenly divisible by 8?
...If yes, N is a square.
...If no, N is not a square.
11--The last two digits of a 3 or more digit square number must be one of the following.
...00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96.
12--The hypotenuse of a primitive Pythagorean triangle is always odd.
The following is a series of general guidelines that will often lead the way to the variety of solutions for varying types of N.
Given: Prime numbers are always of the form 4n + 1 or 4n - 1.
1--Every prime number of the form 4n + 1 is expressable as the sum of two relatively prime squares in one way and is the hypotenuse of a primitive Pythagorean triangle. Both the hypotenuse and its square are both the sum of two squares. A power of the prime or double either the prime or power can also be expressed as the sum of two squares.
2--No prime of the form 4n - 1 can be expressed as the sum of two squares or be the hypotenuse of a primitive Pythagorean triangle.
3--Only composite numbers can be expressed as the sum of two squares in more than one way and only if they contain a certain number of primes of the form 4n + 1.
3--A composite number having only one prime factor of the form 4n + 1 can be expressed as the sum of two squares as long as the other factor, or factors, are powers of 2.
4--A composite number having two different prime factors of the form 4n + 1 can be expressed as the sum of two squares in two ways.
5--A composite number having three different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 4 ways.
6--A composite number having four different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 8 ways.
7--In general, a composite number having k different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 2^(k-1) ways.
8--A composite number having an odd number of prime factors of the form 4n - 1 accompanying at least one prime factor of the form 4n + 1 cannot be expressed as the sum of two squares.
9--A composite number having an even number of prime factors of the form 4n - 1 accompanying at least one prime factor of the form 4n + 1 can be expressed as the sum of two squares, but in only one way. The inclusion of the factor of 2, or any power of 2, will not alter the ability of the number to be expressed as the sum of two Squares nor increase the number of ways that it can be so expressed.
10--If N is a perfect square and the sqrt(N) = z is of the form 4n + 1 (an odd number), z is the hypotenuse of a primitive or non-primitive Pythagorean triangle and N is expressable as the sum of two Squares.
11--If N is a perfect square and the sqrt(N) = z is even, N may be the hypotenuse of a non-primitive Pythagorean triangle.
12--A hypotenuse of a primitive, or non-primitive, Pythagorean triangle must be the form of k(m^2 + n^2).
13--The number N = [2^(a0)xp1^(2a1)xp2^(2a2)xp3^(2a3)x.....pn^(2an)xq1^(b1)xq2^(b2)x.....qr^(br), with p's being primes of the form 4n - 1 and q's being primes of the form 4n + 1, can be exptressed as the sum of two unequal sqares in (b1 + 1)(b2 + 1).....(br + 1)]/2 ways.
14--A positive integer N can be represented as the sum of two squares if, and only if, its factorization into powers of distinct primes contains no odd powers of primes of the form 4n - 1.
15--The product of any two numbers, that are themselves the sums of two squares, can be represented as the sum of two other squares and often in two different ways.
................................= (ac + bd)^2 + (ad - bC)^2
.............................../
(a^2 + b^2)(c^2 + d^2) or
...............................\
...............................= (ac - bd)^2 + (ad + bc)^2
16--If the two numbers happen to be equal to one another, i.e., (a^2 + b^2)(a^2 + b^2), we then have (a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2.
17--In the special case where c = d = 1, 2(a^2 + b^2) = (a + b)^2 + (b - a)^2 meaning that any product with factors of 2 and primes of the form 4n + 1 can be represented as the sum of two squares.
Lets explore some examples of these guidelines.
1--Every prime number of the form 4n + 1 is expressable as the sum of two relatively prime squares in one way and is the hypotenuse of a primitive Pythagorean triangle.
The early primes of the form 4n + 1 are 5, 13, 17, 29, 37, etc.
5 = 2^2 + 1^2, 13 = 3^2 + 2^2, 17 = 4^2 + 1^2, 29 = 5^2 + 2^2, 37 = 6^2 + 1^2, etc.
Both the hypotenuse and its square are the sums of two squares.
5^2 = 25 = 4^2 + 3^2, 13^2 = 169 = 12^2 + 5^2, 17^2 = 289 = 15^2 + 8^2, 29^2 = 841 = 20^2 + 21^2, etc.
A power of the prime or double either the prime or power can be expressed as the sum of two squares.
5^3 = 125 = 11^2 + 2^2, 2x5 = 10 = 3^2 + 1^2, 2x25 = 50 = 7^2 + 1^2, 2x125 = 250 = 15^2 + 5^2.
2--No prime of the form 4n - 1 can be expressed as the sum of two squares or be the hypotenuse of a primitive Pythagorean triangle. Examples of such primes are 3, 7, 11, 19, 23, 31, etc.
3--A composite number having one prime factor of the form 4n + 1 can be expressed as the sum of two squares as long as the other factor, or factors, are powers of 2.
5x2 = 10 = 3^2 + 1^2, 5x4 = 20 = 4^2 + 2^2, 5x8 = 40 = 6^2 + 2^2, 13x2 = 26 = 5^2 + 1^2, 13x4 = 52 = 6^2 + 4^2, 13x8 = 104 = 10^2 + 2^2, etc.
4--A composite number having two different prime factors of the form 4n + 1 can be expressed as the sum of two squares in two ways. Since primes of the form 4n + 1 are expressable as the sum of two squares to begin with, the product of two such primes are expressable as the sum of two squares in two ways according to the relationship (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 and (ac - bd)^2 + (ad + bc)^2.
65 = 5x13 = (2^2 + 1^2)(3^2 + 2^2) = (8^2 + 1^2) and (7^2 + 4^2).
85 = 17x5 = (4^2 + 1^2)(2^2 + 1^2) = (9^2 + 2^2) and (7^2 + 6^2).
221 = 13x17 = (3^2 + 2^2)(4^2 + 1^2) = (14^2 + 5^2) and (11^2 + 10^2).
5--A composite number having three different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 4 different ways in accordance with the same relationship described in (4).
1105 = 5x13x17 = 65x17 = (8^2 + 1^2)(4^2 + 1^2) = (33^2+ 4^2) = (31^2 + 12^2)
.....................................= (7^2 + 4^2)(4^2 + 1^2) = (32^2 + 9^2) = (24^2 + 23^2)
...........................85x13 = (9^2 + 2^2)(3^2 + 2^2) = (31^2 + 12^2) = 23^2 + 24^2)
.....................................= (7^2 + 6^2)(3^2 + 2^2) = (33^2 + 4^2) = (9^2 + 32^2)
...........................221x5 = (14^2 + 5^2)(2^1 + 1^2) = (33^2 + 4^2) = (23^2 + 24^2)
.....................................= (11^2 + 10^2)(2^2 + 1^2) = (32^2 + 9^2) = 12^2 + 31^2)
The highlighted solutions are the four unique solutions which derive from either of the three combinations of the three factors.
6--A composite number having four different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 8 ways.
7--In general, a composite number having k different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 2^(k-1) ways.
8--A composite number having an odd number of prime factors of the form 4n - 1 cannot be expressed as the sum of two squares.
3x5 = 15, 3x9 = 21, 7x17, 19x29, 7x7x7x13, 11x11x11x5, etc. will not work.
9--A composite number having an even number of prime factors of the form 4n - 1 an be expressed as the sum of two squares, but in only one way.
3x3x13 = 117 = (9^2 + 6^2), 7x7x17 = 833 = (28^2 + 7^2).
10--If N is a perfect square and the sqrt(N) = z is of the form 4n + 1, z is the hypotenuse of a Pythagorean triangle and N is expressable as the sum of two Squares.
sqrt(25) = 5 = (2^2 + 1^2), sqrt(169) = 13 = (3^2 + 2^2), etc.
11--If N is a perfect square and the sqrt(N) = z is even, N may be the hypotenuse of a non-primitive Pythagorean triangle.
With the above material as general background to the subject, we can now generally state that the solution of x^2 + y^2 = N, N being composite, derive from the specific theorems stated earlier:
1) A positive integer N can be represented as the sum of two squares if, and only if, its factorization into powers of distinct primes contains no odd powers of primes of the form 4n - 1.
2) The product of any two numbers, that are themselves the sums of two squares, can be represented as the sum of two other squares and often in two different ways.
................................= (ac + bd)^2 + (ad - bC)^2
.............................../
(a^2 + b^2)(c^2 + d^2) or
...............................\
...............................= (ac - bd)^2 + (ad + bc)^2
3) If the two numbers happen to be equal to one another, i.e., (a^2 + b^2)(a^2 + b^2), we then have (a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2.
4) In the special case where c = d = 1, 2(a^2 + b^2) = (a + b)^2 + (b - a)^2 meaning that any product with factors of 2 and primes of the form 4n + 1 can be represented as the sum of two squares.
Now the moment we have been waiting for. How do we make use of all of this information to solve problems of this type x^2 + y^2 = N?
Out of the first 25 primes, 2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,
and 97, the following are of the form 4n + 1; 5, 13, 17, 29, 37, 41, 53, 61, 73, 79, 89, 97. The following are of the
form 4n - 1; 3, 7, 19, 23, 31, 43, 59, 67, 71, 79, and 89.
Observe that according to the rules established earlier, 5 = (2^2+ 1^2), 13 = (3^2 + 2^2), 17 = (4^2 + 1^2), 29 = (5^2+2^2), 37 = (6^2+ 1^2), 41 = (5^2+ 4^2), and 53 = (7^2+2^2), etc.
Consider a simple example of x^2 + y^2 = 65.
The prime factorization of 65 is 5^2(13^1).
These primes can be represented as the sum of two squares by 5 = (2^2 + 1^2) and 13 = (3^2 + 2^2).
Therefore,
................................ ............a......b.....c.....d
...................... 65 = 5x13 = (2^2+ 1^2)(3^2+2^2) with a, b, c, and d equaling 3, 1, 5, and 2.
From our identities,
(2^2+ 1^2)(3^2+2^2) = (2x3 + lx2)^2 + (2x2 - 1x3)^2 = (6 + 2)^2 + (4 - 3)^2 = (8^2 + 1^2) = 64 + 1 = 65
.....................................................................and
(2^2+ 1^2)(3^2+2^2) = (2x3 - 1x2)^2 + (2x2 + 1x3)^2 = (6 - 2)^2 + (4+3)^2 = (4^2 + 7^2) = 16 + 49 = 65.
Therefore, 65 = (8^2 + 1^2) or (4^2+7^2).
Consider N = x^2 + y^2 = 377.
.........................a.......b.....c......d
377 = 13x29 = (3^2 + 2^2)(5^2 + 2^2) with a, b, c, and d equaling 3, 2, 5, and 2, we can make use of
the identities to create
(3^2+2^2)^2 + 2^2) = (3x5 + 2x2)^2 + (3x2 - 2x5)^2 = (15 + 4)^2 + (6 -10)^2 = (19^2 + 4^2) = 361 + 16 and
.......................... = (3x5 - 2x2)^2 + (3x2 + 2x5)^2 = (15 - 4)^2 + (6 + 10)^2 = (11^2+16^2) = 121 + 256 = 377.
Therefore, 377 = (19^2 + 4^2) or (11^2 + 16^2).
Consider N = x^2 + y^2 = 315.
The prime factors of this number are = 3^2(5^1)7^1. Since 315 is of the form 4n - 1 and the factorization contains an odd power of the prime 7 of the form 4n - 1, it cannot be represented as the sum of two squares.
Consider the number x^2 + y^2 = 3185.
The prime factors of 3185 are 5^1(7^2)13^1. Since 3185 is of the form 4n + 1, and/or the factorization contains no odd power of a prime of the form 4n - 1, (7^2 has an even power), the number is representable as a sum of two squares. Having 5 = (2^2+ 1^2), 7^2 = (7^2+ 0^2), and 13 = (3^2 + 2^2), we can write
3185 = 5x(7^2)x13 = (2^2 + 1^2)(7^2 + 0^2)(3^2 + 2^2)
........= (14^2 + 7^2)(3^2 + 2^2)
........= (14x3 + 7x2)^2 + (14x2 - 7x3)^2 or (14x3 - 7x2)^2 + (14x2 + 7x3)^2
........= (42 + 14)^2 + (28 - 21)^2 or (42 - 14)^2 + (28 + 21)^2
........= (56^2 + 7^2) or (28^2 + 49^2)
........= (3136 + 49) = 3185 or (784 + 2401) = 3185.
So far, we can see that the solutions are relatively straight forward and simple for small numbers. Lets see how well we can apply the method to a large number.
Consider x^2 + y^2 = 1,395,182,880
The prime factors of this number are 2^5(3^4)5^1(7^2)13^3.
Being of the form 4n + 1, both 5 and 13 are the sums of two squares.
We already know that 5x13 = 65 can be written as (8^2 + 1^2) and (4^2 + 7^2)
We can rewrite our factorization of 2^5(3^4)5^1(7^2)13^3 as (2x5x13)x[2^4(3^4)7^2(13^2)].
From our third identity above, 2(5x13) = 9^2 + 7^2 and 11^2 + 3^2.
We can also rewrite [2^4(3^4)7^2(13^2) as [2^2(3^2)7^1(13^1)]^2
Then, 2^2(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)9^1]^2 + [2^2(3^2)7^1(13^1)7^1]^2
......................................= 29,484^2 + 22,932^2
.......................................= 869,306,256 + 525,876,624
.......................................= 1,395,182,880
..................................................and
........2^5(^3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)11^1] + [2^2(3^2)7^1(13^1)3^1]
....................................... = 36,036^2 + 9,828^2
.......................................= 1,298,593,296 + 96,589,584
.......................................= 1,395,182,880
From statement 13 above, we can see that the number 2^5(^3^4)5^1(7^2)13^3 can be expressed as the sum of two squares in (1 + 1)(3 + 1)/2 = 4 different ways.
Thus, we can rewrite our factorization of 2^5(3^4)5^1(7^2)13^3 as (2x5x13^3)x[2^4(3^4)7^2].
(2x5x13^3) can be written as 2(104^2 + 13^2) and 2(101^2 + 28^2).
From our third identity above, 2x5x13^3 = (117^2 + 91^2) and (129^2 + 73^2).
We can also rewrite [2^4(3^4)7^2) as [2^2(3^2)7^1]^2.
Then, 2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(117^1)]^2 + [2^2(3^2)7^1(91)]^2
.......................................= 29484^2 + 22932^2
.......................................= 869,306,256 + 525,876,624
.......................................= 1,395,182,880.
..................................................and
.........2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(129)]^2 + [2^2(3^2)7^1(73)]^2
.......................................= 32508^2 + 18396^2
.......................................= 1,056,770,064 + 338,412,816
.......................................= 1,395,182,880
Consider N = x^2 + y^2 = 7,429,968,000.
The prime factors of this number are 2^7(3^6)5^3(7^2)10 3(13^1)
This factorization can be rewritten as (2x5x13)[2^6(3^6)5^2(7^2)]
We know from above that 5xl 3 = (8^2 + 1^2) and (4^2 + 7^2)
We can then express 2(5x13) by (9^2 + 7^2 and (11^2 + 3^2).
We can rewrite 2^6(3^6)5^2(7^2) as [2^3(3^3)5^1(7^1)]^2
Having 10xl 3 = 11^2 + 3^2 and 9^2 + 7^2 we can now write
[2^7(3^6)5^3(7^2)13^1] = [2^3(3^3)5^1(7^1)11^1]^2 + [2^3(3^3)5^1(7^1)3^1]^2
............7,429,968,000 = 83,160^2 + 22,680^2
.................................= 6,915,585,600 + 514,382,400
.................................= 7,429,968,000
and
[2^4(3^6)7^2(10^3)13^1] = [2^3(3^3)5^1(7^1)9^1]^2 + [2^3(3^3)5^1(7^1)7^1]^2
.............7,429,968,000 = 68,040^2 + 52,920^2
................................. = 4,629,441,600 + 2,800,526,400
..................................= 7,429,968,000
I'll let you figure out the other two ways to express this number as the sum of two squares.
* There are an infinite number of rational solutions to x^2 + y^2 = 1. Assuming any rational value of m, x = 2m/(m^2 + 1) and y = (1 - m^2)/(1 + m^2)
Oh, I see you've discovered the secret formula for perfect numbers! It's quite simple, really. The only perfect number of the form x^n + y^n is... drumroll, please... well, actually, there isn't one! Perfect numbers are a bit of a tricky bunch and don't follow that specific pattern. They're more like that one friend who always surprises you with something unexpected. So, keep searching for that perfect number, but don't worry if you can't find it with that formula. Just remember, math can be a bit unpredictable, just like life!
The concept of a perfect number does not involve exponents. A perfect number is a positive integer that is equal to the sum of its positive divisors excluding itself. The only known even perfect numbers are of the form (2^(p-1))*(2^p - 1), where 2^p-1 is a prime number. So, a perfect number cannot be expressed as x^n + y^n.
To determine if there exists a perfect number of the form x^n + y^n, we need to analyze the properties of perfect numbers and investigate whether such numbers can be expressed in that particular form.
A perfect number is a positive integer that is equal to the sum of its proper divisors (excluding the number itself). In other words, if we add up all the numbers that divide evenly into a perfect number, except the number itself, the sum will be equal to the original number.
The equation x^n + y^n suggests that we are working with two variables, x and y, raised to the power of n. By substituting different values of x, y, and n, we can check if the resulting number is a perfect number.
Here's one approach to find a perfect number of the form x^n + y^n:
1. Choose a value for n. Start with a small value, such as 2.
2. Select values for x and y. They can be any positive integers.
3. Calculate x^n + y^n.
4. Check if the result is a perfect number by summing up all the proper divisors of that result.
- If the sum of the proper divisors is equal to the result, then it is a perfect number.
- If not, try different values for x and y, or increase the value of n and repeat steps 3-4.
Note: Finding a perfect number in the form of x^n + y^n can be challenging, as there is no guarantee that such a number exists. Additionally, for larger values of n, x, and y, it becomes computationally more demanding to search for potential solutions.
It is worth mentioning that the search for perfect numbers has been an ongoing area of study in mathematics, and as of now, only a few perfect numbers have been discovered.