Fermat's Two Squares Theorem

[jillbones]

Fermat's Two Squares Theorem was originally stated as; "Prime numbers may be expressed as the sum of two squares." Subsequent investigation revealed this was true only if P = 4*k + 1. " I have developed a simple proof for this conjecture. As follows ...

Given: P= 4*k + 1 = 4 *m + 4 * (k - m) + 1. I can show that every even square is equal to 4*n^2 and every odd square is equal to
4*(x^2 + x) + 1.

Discussion:
The equation "4*k +1 = 4 *m + 4 * (k - m) + 1" is a mathematical identity.
For any prime P all squares less than P can be generated by this equation.

These two facts should be sufficient for proof of the theorem but I am uncertain.

Also, I believe that the theorem should be expanded to include all numbers of the form 4*k +1.

 

[JeffM]

What you have written is virtually unintelligible.

First, what are you trying to prove: Fermat's original conjecture or the narrower modern statement.

Second, I suspect that what you are trying to say in your second paragraph is

[MATH]m \text { is an even perfect square} > 0 \implies \\ \exists \text { positive integer } n \text { such that } m = 4n^2.\\ m \text { is an odd perfect square} \implies \\ \exists \text { non-negative integer } n \text { such that } m = (2n + 1)^2 = 4n^2 + 4n + 1.[/MATH]That is true (although the way you stated the odd case is not quite consistent with traditional number theory). If you are trying to say something else, please clarify.

Third, you are correct:

[MATH]4u + 1 \equiv 4v - 4v + 4u + 1 \equiv 4v + 4u - 4v + 1 \equiv 4v + 4(u - v) + 1.[/MATH]
You then say "For any prime P, all [perfect] squares less than P can be generated by this formula."

This is complete and utter nonsense. Let's take P = 29. How do you generate 2 and 16, which are perfect squares less than the prime 29, by this formula?

Fourth, I do not want to be snarky. All primes > 2 are odd. So perhaps what you meant to say is "For any prime P, all [odd, perfect] squares less than P can be generated by this formula." That is true. Let P = 83.

9 = 4 * 2 + 1

25 = 4 * 6 + 1

49 = 4 * 12 + 1

81 = 4 * 20 + 1

Odd perfect squares less than the prime 83 can be expressed as 4k + 1. Lovely!

But of course by definition a prime IS NOT A PERFECT SQUARE.

So the whole process is completely irrelevant.

Number theory is hard.

 

[Cubist]

Also, I believe that the theorem should be expanded to include all numbers of the form 4*k +1.

In this last part, are you saying that 9, 21 or 33 (etc) can be made from the sum of two (non-zero) squares?

--

Let set A be the set of numbers generated by 4*n^2 + 4*(x^2 + x) + 1 where n and x are integers an n>0 and x>=0
set B is generated by 4*k + 1 where k is an integer and k>0
set C is the intersection of B and the set of prime numbers

I think you've proved that the set of numbers A are all contained in the set of numbers B

...but you haven't proved that every element of C is within A and I think this would be required for a proof

 

[Steven G]

Nothing against you at all but you are not going to come up with an easier proof. Sorry!

There are research mathematicians who specialize in number theory, some even dedicating their entire career to number theory and they would have seen an easier proof by now.

 

[JeffM]

Let’s formulate the proposition to be proved.

Every odd number can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer. Every prime number > 2 is odd. Therefore, every prime number > 2 can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer.

Consider 11. It is a prime > 2. It is equal to 4 * 2 + 3. The only perfect squares less than 11 are 1, 4, and 9.
1+ 4 = 5, 1 + 9 = 10, 4 + 9 = 13. 11 is a prime not expressible as the sum of two perfect square.

Consider 13. It too is a prime > 2. It is equal to 4 * 3 + 1. It equals 2 squared + 3 squared.

Proposition to be proved.

If p is any prime expressible as 4k + 1, where k is a positive integer, then there exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
If p is any prime that is not expressible as 4k + 1, where k is a positive integer, then there do not exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
Now you have something to sink your teeth into.

 

[jillbones]

Let me explain the equation; "4*m + 4*(k-m) + 1 = 4*k + 1. 4*m is the square of an even number. 4*(k-m)+ 1 is the
square of an odd number. 4*k+1 is an integer sometimes a prime integer. The equation infers that some primes the sum of an even integer and an odd integer. It also says that 0 = 0, which makes it valid for all 'k' and all 'm' .

Consider primes of the form 4*k + 3 = 4*m + 4*(k-m) + 3. Now 4*(k-m) + 3 is an odd square. But 4*(k-m) + 3
cannot be a square if 4*(k-m) + 1 is.

 

[jillbones]

Nothing against you at all but you are not going to come up with an easier proof. Sorry!

There are research mathematicians who specialize in number theory, some even dedicating their entire career to number theory and they would have seen an easier proof by now.

Look at it as algebra problem!

 

[jillbones]

Let’s formulate the proposition to be proved.

Every odd number can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer. Every prime number > 2 is odd. Therefore, every prime number > 2 can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer.

Consider 11. It is a prime > 2. It is equal to 4 * 2 + 3. The only perfect squares less than 11 are 1, 4, and 9.
1+ 4 = 5, 1 + 9 = 10, 4 + 9 = 13. 11 is a prime not expressible as the sum of two perfect square.

Consider 13. It too is a prime > 2. It is equal to 4 * 3 + 1. It equals 2 squared + 3 squared.

Proposition to be proved.

If p is any prime expressible as 4k + 1, where k is a positive integer, then there exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
If p is any prime that is not expressible as 4k + 1, where k is a positive integer, then there do not exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
Now you have something to sink your teeth into.

How have I failed to accomplish this?

 

[JeffM]

Let me explain the equation; "4*m + 4*(k-m) + 1 = 4*k + 1. 4*m is the square of an even number. 4*(k-m)+ 1 is the
square of an odd number. 4*k+1 is an integer sometimes a prime integer. The equation infers that some primes the sum of an even integer and an odd integer. It also says that 0 = 0, which makes it valid for all 'k' and all 'm' .

Consider primes of the form 4*k + 3 = 4*m + 4*(k-m) + 3. Now 4*(k-m) + 3 is an odd square. But 4*(k-m) + 3
cannot be a square if 4*(k-m) + 1 is.

No. 4 * 8 = 32, and there is no integer, which when squared equals 32. You are so far off base that, to quote Fermi, “you are not even wrong.” 4 * 8 + 1 = 33, which is not prime.

Try reading what was previously posted and try to understand it. We shall answer questions about meaning, but will conclude you are a troll if ignore what was posted.

 

[jillbones]

In this last part, are you saying that 9, 21 or 33 (etc) can be made from the sum of two (non-zero) squares?

--

Let set A be the set of numbers generated by 4*n^2 + 4*(x^2 + x) + 1 where n and x are integers an n>0 and x>=0
set B is generated by 4*k + 1 where k is an integer and k>0
set C is the intersection of B and the set of prime numbers

I think you've proved that the set of numbers A are all contained in the set of numbers B

...but you haven't proved that every element of C is within A and I think this would be required for a proof

9 , 21 & 33 are not primes. I apologize if I inferred this.

 

[JeffM]

9 , 21 & 33 are not primes. I apologize if I inferred this.

Try saying what you mean exactly.

 

[Cubist]

OP, I think it's good that you're trying to find an alternative proof. Even if you don't succeed you'll very probably improve your mathematical reasoning skills. But don't get your hopes up because the odds are stacked against you for reasons stated above!

How have I failed to accomplish this?

I'll try to explain in a different way. You wrote...

P= 4*k + 1 = 4 *m + 4 * (k - m) + 1

but there's a logical error in the above because it doesn't accurately reflect the original problem statement.

--

Start by precisely defining your variables, any constraints, and be more explicit about the thing that you're trying to prove.

It's better to write...

GIVEN P=4*k + 1
where P is also an element of the set of prime numbers, and k is an integer where k>0

(the above says that P is an element of a subset of the prime numbers)

PROVE for every possible value of P, that there exists a value of m such that...

• P = 4 *m + 4 * (k - m) + 1
• AND m is an integer
• AND m≥1
• AND 4*m = 4*n^2 where n is an integer
• AND 4*(x^2 + x) + 1 = 4 * (k - m) + 1 where x is an integer and x≥0

This better reflects the actual problem (and I think it reflects what you were aiming for). I think you probably understand these requirements from what you've already written. But because you wrote your OP very quickly, not not very clearly, I think you skipped the part where you have to prove that all constraints are satisfied simultaneously.

NOTE: It isn't a proof if you simply say, "all the constraints are satisfied by some value of m". Why? Which value of m? All the "=" signs after the word "PROVE" are not necessarily true until you can prove that they are.

 

[jillbones]

Let’s formulate the proposition to be proved.

Every odd number can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer. Every prime number > 2 is odd. Therefore, every prime number > 2 can be expressed as 4k + 1 or as 4k + 3 where k is a non-negative integer.

Consider 11. It is a prime > 2. It is equal to 4 * 2 + 3. The only perfect squares less than 11 are 1, 4, and 9.
1+ 4 = 5, 1 + 9 = 10, 4 + 9 = 13. 11 is a prime not expressible as the sum of two perfect square.

Consider 13. It too is a prime > 2. It is equal to 4 * 3 + 1. It equals 2 squared + 3 squared.

Proposition to be proved.

If p is any prime expressible as 4k + 1, where k is a positive integer, then there exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
If p is any prime that is not expressible as 4k + 1, where k is a positive integer, then there do not exist positive integers q and r such that

[MATH]q^2 + r^2 = p.[/MATH]
Now you have something to sink your teeth into.

Given 4*k +1 = q^2 + r^2 = p. Let "4*m = q^2" & and "4*(k-m) +1 = r^2". This is always true if m = (q^2)/4; and (k-m) = (r^2-1)/4. Observation: All even squares are mod4. All odd squares less one are also mod4!
This confirms the first part of the proposition.

If 4*k-1 = 4*m +4*(k-m) - 1, it is clear that 4*(k-m) - 1 must be a perfect square. This is impossible if
4*(k-m)+1 is also a square. This confirms the second part of the proposition.

I consider this problem to be merely an exercise in algebra. Perhaps if the ancients had thought in this way;
they would have not missed the answer. But If you're playing with primes and number theory - whose got time
for poor little algebra

 

[JeffM]

You are assuming what is to be proved.

It is not universally true that 4 times a positive integer plus 1 equals the sum of two perfect squares. Suppose k = 5. Then 4k + 1 = 21. There are four perfect squares < 21, there are eight sums to consider.

1 + 1 = 2, not 21.

1 + 4 = 5, not 21.

1 + 9 = 10, not 21.

1 + 16 = 17, not 21.

4 + 4 = 8, not 21.

4 + 9 = 13, not 21.

4 + 16 = 20, not 21.

9 + 9 = 18, not 21.

It is of course easy to prove something if you start by assuming a falsity.

Some integers expressible as 4k + 1 do equal the sum of two perfect squares (13 for example), and some do not (21 for example).

Some primes are expressible as the sum of 4k + 1 (5 for example), and some are not (7 for example).

You start by assuming a number that is in the intersection of those sets and then prove that number must be in the intersection of those sets. Brilliant: the Fields Prize awaits.

Now it is obvious that the intersection of the primes expressible as 4k + 1 and the integers that are expressible as 4k + 1 and as the sum of two perfect squares is not empty, 13 is an example. But you have come nowhere close to showing that every prime of that type is one of the integers expressible as the sum of two perfect squares. Your whole proof is premised on the assumption that all integers expressible as 4k + 1 can also be expressed as the sum of two perfect squares, but that is false.

You have given no evidence that you even understand the proposition that you claim to have proved.

 

[jillbones]

You are assuming what is to be proved.

It is not universally true that 4 times a positive integer plus 1 equals the sum of two perfect squares. Suppose k = 5. Then 4k + 1 = 21. There are four perfect squares < 21, there are eight sums to consider.

1 + 1 = 2, not 21.

1 + 4 = 5, not 21.

1 + 9 = 10, not 21.

1 + 16 = 17, not 21.

4 + 4 = 8, not 21.

4 + 9 = 13, not 21.

4 + 16 = 20, not 21.

9 + 9 = 18, not 21.

It is of course easy to prove something if you start by assuming a falsity.

Some integers expressible as 4k + 1 do equal the sum of two perfect squares (13 for example), and some do not (21 for example).

Some primes are expressible as the sum of 4k + 1 (5 for example), and some are not (7 for example).

You start by assuming a number that is in the intersection of those sets and then prove that number must be in the intersection of those sets. Brilliant: the Fields Prize awaits.

Now it is obvious that the intersection of the primes expressible as 4k + 1 and the integers that are expressible as 4k + 1 and as the sum of two perfect squares is not empty, 13 is an example. But you have come nowhere close to showing that every prime of that type is one of the integers expressible as the sum of two perfect squares. Your whole proof is premised on the assumption that all integers expressible as 4k + 1 can also be expressed as the sum of two perfect squares, but that is false.

You have given no evidence that you even understand the proposition that you claim to have proved.

The proposition says that PRIMES = 4*k +1 can be expressed as the sum of two squares and those primes = 4*k-1
cannot.

OP, I think it's good that you're trying to find an alternative proof. Even if you don't succeed you'll very probably improve your mathematical reasoning skills. But don't get your hopes up because the odds are stacked against you for reasons stated above!



I'll try to explain in a different way. You wrote...

P= 4*k + 1 = 4 *m + 4 * (k - m) + 1

but there's a logical error in the above because it doesn't accurately reflect the original problem statement.

--

Start by precisely defining your variables, any constraints, and be more explicit about the thing that you're trying to prove.

It's better to write...

GIVEN P=4*k + 1
where P is also an element of the set of prime numbers, and k is an integer where k>0

(the above says that P is an element of a subset of the prime numbers)

PROVE for every possible value of P, that there exists a value of m such that...

• P = 4 *m + 4 * (k - m) + 1
• AND m is an integer
• AND m≥1
• AND 4*m = 4*n^2 where n is an integer
• AND 4*(x^2 + x) + 1 = 4 * (k - m) + 1 where x is an integer and x≥0

This better reflects the actual problem (and I think it reflects what you were aiming for). I think you probably understand these requirements from what you've already written. But because you wrote your OP very quickly, not not very clearly, I think you skipped the part where you have to prove that all constraints are satisfied simultaneously.

NOTE: It isn't a proof if you simply say, "all the constraints are satisfied by some value of m". Why? Which value of m? All the "=" signs after the word "PROVE" are not necessarily true until you can prove that they are.

In this last part, are you saying that 9, 21 or 33 (etc) can be made from the sum of two (non-zero) squares?

--

Let set A be the set of numbers generated by 4*n^2 + 4*(x^2 + x) + 1 where n and x are integers an n>0 and x>=0
set B is generated by 4*k + 1 where k is an integer and k>0
set C is the intersection of B and the set of prime numbers

I think you've proved that the set of numbers A are all contained in the set of numbers B

...but you haven't proved that every element of C is within A and I think this would be required for a proof

In this last part, are you saying that 9, 21 or 33 (etc) can be made from the sum of two (non-zero) squares?

--

Let set A be the set of numbers generated by 4*n^2 + 4*(x^2 + x) + 1 where n and x are integers an n>0 and x>=0
set B is generated by 4*k + 1 where k is an integer and k>0
set C is the intersection of B and the set of prime numbers



I think you've proved that the set of numbers A are all contained in the set of numbers B

...but you haven't proved that every element of C is within A and I think this would be required for a proof

In this last part, are you saying that 9, 21 or 33 (etc) can be made from the sum of two (non-zero) squares?

--

Let set A be the set of numbers generated by 4*n^2 + 4*(x^2 + x) + 1 where n and x are integers an n>0 and x>=0
set B is generated by 4*k + 1 where k is an integer and k>0
set C is the intersection of B and the set of prime numbers

I think you've proved that the set of numbers A are all contained in the set of numbers B

...but you haven't proved that every element of C is within A and I think this would be required for a proof

NO Actually, the set of numbers B are all contained in the set of numbers A. Set A includes most of the
composite numbers as well as those prime numbers included in set B]. Set B includes only prime numbers.

Suppose p = 4*k -1. Then 4*k - 1 = 4*m + 4*(k-m) -1. Then 4*(k-m) -1 is perfect square. If 4*(k-m) + 1 is a perfect square; 4*(k-m) -1 cannot be a perfect square.
.

 

[JeffM]

In the spirit of cubist's most recent post.

First, the standard way to show odd numbers in number theory is

[MATH][/MATH]

The proposition says that PRIMES = 4*k +1 can be expressed as the sum of two squares and those primes = 4*k-1
cannot.

Great. You do understand the proposition. I apologize for the snark.

I agree that some (but not all) odd numbers of the form 4k + 1 can be expressed as the sum of two perfect squares.

I agree that all prime numbers > 2 are odd.

I further agree that some odd prime numbers (but not all) can be expressed as both equal to 4k + 1 and as the sum of two perfect squares.

And finally I DO AGREE that if, contrary to fact, all odd numbers expressible as 4k + 1 were also expressible as the sum of two perfect squares, then it would be trivial to show that all primes expressible as 4k + 1 are also expressible as the sum of two perfect squares. But the premise is false.

The challenge is to prove that (1) all odd numbers expressible as 4k + 1 and not expressible as the sum of two perfect squares are not prime, and (2) every prime expressible as 4k - 1 is not expressible as the sum of two squares.

You understand the proposition, but you do not understand what it takes to prove it.

Lastly, I do not know where you got your information about the the theorem. Fermat was not an ancient. He was a contemporary of the French mathematicians who put algebra in its modern form and lived almost 5000 years after the Sumerian mathematicians. Nor could Fermat have possibly have asserted that every prime is the sum of two perfect squares. It is false for three, seven, and eleven, three out of four of the first four odd primes. Fermat may have been loose about proofs, but he was not an idiot.

I wish you good luck on finding a simple proof. That would show a much, much greater talent for mathematics than I can aspire to. But please understand that your simple alleged proof is founded on an error.

 

[jillbones]

In the spirit of cubist's most recent post.

First, the standard way to show odd numbers in number theory is

[MATH][/MATH]
Great. You do understand the proposition. I apologize for the snark.

I agree that some (but not all) odd numbers of the form 4k + 1 can be expressed as the sum of two perfect squares.

I agree that all prime numbers > 2 are odd.

I further agree that some odd prime numbers (but not all) can be expressed as both equal to 4k + 1 and as the sum of two perfect squares.

And finally I DO AGREE that if, contrary to fact, all odd numbers expressible as 4k + 1 were also expressible as the sum of two perfect squares, then it would be trivial to show that all primes expressible as 4k + 1 are also expressible as the sum of two perfect squares. But the premise is false.

The challenge is to prove that (1) all odd numbers expressible as 4k + 1 and not expressible as the sum of two perfect squares are not prime, and (2) every prime expressible as 4k - 1 is not expressible as the sum of two squares.

You understand the proposition, but you do not understand what it takes to prove it.

Lastly, I do not know where you got your information about the the theorem. Fermat was not an ancient. He was a contemporary of the French mathematicians who put algebra in its modern form and lived almost 5000 years after the Sumerian mathematicians. Nor could Fermat have possibly have asserted that every prime is the sum of two perfect squares. It is false for three, seven, and eleven, three out of four of the first four odd primes. Fermat may have been loose about proofs, but he was not an idiot.

I wish you good luck on finding a simple proof. That would show a much, much greater talent for mathematics than I can aspire to. But please understand that your simple alleged proof is founded on an error.

 

[jillbones]

In the spirit of cubist's most recent post.

First, the standard way to show odd numbers in number theory is

what?

[MATH][/MATH]
Great. You do understand the proposition. I apologize for the snark.

I agree that some (but not all) odd numbers of the form 4k + 1 can be expressed as the sum of two perfect squares.

I agree that all prime numbers > 2 are odd.

I further agree that some odd prime numbers (but not all) can be expressed as both equal to 4k + 1 and as the sum of two perfect squares.

And finally I DO AGREE that if, contrary to fact, all odd numbers expressible as 4k + 1 were also expressible as the sum of two perfect squares, then it would be trivial to show that all primes expressible as 4k + 1 are also expressible as the sum of two perfect squares. But the premise is false.

The challenge is to prove that (1) all odd numbers expressible as 4k + 1 and not expressible as the sum of two perfect squares are not prime, and (2) every prime expressible as 4k - 1 is not expressible as the sum of two squares.

You understand the proposition, but you do not understand what it takes to prove it.

Lastly, I do not know where you got your information about the the theorem. Fermat was not an ancient. He was a contemporary of the French mathematicians who put algebra in its modern form and lived almost 5000 years after the Sumerian mathematicians. Nor could Fermat have possibly have asserted that every prime is the sum of two perfect squares. It is false for three, seven, and eleven, three out of four of the first four odd primes. Fermat may have been loose about proofs, but he was not an idiot.

I wish you good luck on finding a simple proof. That would show a much, much greater talent for mathematics than I can aspire to. But please understand that your simple alleged proof is founded on an error.

what?

Challenge (3) Prove that all primes expressible as 4k + 1 are expressible as the sum of two perfect squares.