# Conjecture re form x^2 + Mxy + y^2

• ramsey2879
In summary: Please provide a proof?In summary, the conversation discusses a conjecture that states if x and y are coprime and M does not equal 2, then the equation x^2 + Mxy + y^2 = p^2 has integral solutions only for p = a prime or for products of such primes. It is also mentioned that for M > 2, if x and y are solutions, then y and (My - x) are also solutions. However, this conjecture may not hold true for certain primes such as 3, 5, 11, etc. It is also noted that if n is a product of primes, then each of the primes and their powers and products can be expressed in
ramsey2879
Conjecture: If x and y are coprime and M <> 2 then x^2 + Mxy +y^2 = p^2 has integral solutions only for p = a prime or for products of such primes. Also if M is positive then
both x and y are partial solutions for X^2 -MXY + Y^2 = p^2. Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49

x^2 +Mxy +y^2 == 0 mod n

for M<2, x^2 +Mxy +y^2 = (x+y)^2 - kxy, where M = 2-k

for M>2, x^2 +Mxy +y^2 = (x+y)^2 + kxy, where M = 2+k

$$(x+y)^2 == \pm kxy \ mod \ n$$

as n do not divide (x+y), then for euler

$$(x+y)^{\varphi(n)} == 1 \ mod \ n$$

supose n is composite ==> $$\varphi(n)} = 2^ah$$ such that a > 1 and h is odd ==>

==> $$\huge (x+y)^{2^{2^{a-1}h}} = (x+y)^{2^ah} = (x+y)^{\varphi(n)} == (\pm kxy)^{2^{a-1}h} == 1 \ mod \ n$$

call $$(\pm kxy)^{2^{a-1}} = w$$, and w is obviously positive, then

$$w^h == 1 \ mod \ n$$, an absurdum

**then n is prime**

al-mahed said:
x^2 +Mxy +y^2 == 0 mod n

for M<2, x^2 +Mxy +y^2 = (x+y)^2 - kxy, where M = 2-k

for M>2, x^2 +Mxy +y^2 = (x+y)^2 + kxy, where M = 2+k

$$(x+y)^2 == \pm kxy \ mod \ n$$

as n do not divide (x+y), then for euler

$$(x+y)^{\varphi(n)} == 1 \ mod \ n$$

supose n is composite ==> $$\varphi(n)} = 2^ah$$ such that a > 1 and h is odd ==>

==> $$\huge (x+y)^{2^{2^{a-1}h}} = (x+y)^{2^ah} = (x+y)^{\varphi(n)} == (\pm kxy)^{2^{a-1}h} == 1 \ mod \ n$$

call $$(\pm kxy)^{2^{a-1}} = w$$, and w is obviously positive, then

$$w^h == 1 \ mod \ n$$, an absurdum

**then n is prime**

But for n = 13*7 and M = 1 there is a solution, , i.e. x = 39, y = 65, how does that agree with your conclusion that n must be prime?

Nevermind, I made a mistake in the proof, the conclusion is wrong, I tried to repair the proof without success.

Please, define a little more clear your conjecture, what you mean saying "integral solutions"?

ramsey2879 said:
Conjecture: If x and y are coprime and M <> 2 then x^2 + Mxy +y^2 = p^2 has integral solutions only for p = a prime or for products of such primes. Also if M is positive then
both x and y are partial solutions for X^2 -MXY + Y^2 = p^2. Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49
Another example
7*7 + 7*8 + 8*8 = 13^2

19^2 + 19*80 + 80^2 = 49^2*13^2 = (7*13)^2

Please disreegard the solution in my last post since 39 and 65 are not coprime.
By integral solutions, I mean coprime positive integers.

in fact n = 13*7*91, your conjecture is about p and p^2 only? or what? every composite number? I did not follow this part.

ok, so the conjeture is: for x and y coprime, the expression is equal to some power of a prime number OR some power or a semi-prime number (a product with no more than 2 different factors)?

if your conjecture is about only squares, then we can use the fact that a square is a number of the form 8n+1 where n is a triangular number

Iff M=1 and (x,y,p) are positive coprime integers, then p = a prime of the form 6n+1, or a multiple thereof, or a product of two or more primes thereof.

Similar to Pythagorean Triples where the equation is x^2+y^2=p^2 and p = a prime of the form 4n+1...

Added a bit later... Alternatively, for every prime p, of the form 6n+1, there is one and only one pair of positive integers (x,y) that satisfy the equation x^2+xy+y^2=p^2 (where x and y are not equal and are interchangeable).

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Ynaught? said:

Iff M=1 and (x,y,p) are positive coprime integers, then p = a prime of the form 6n+1, or a multiple thereof, or a product of two or more primes thereof.

Similar to Pythagorean Triples where the equation is x^2+y^2=p^2 and p = a prime of the form 4n+1...

Added a bit later... Alternatively, for every prime p, of the form 6n+1, there is one and only one pair of positive integers (x,y) that satisfy the equation x^2+xy+y^2=p^2 (where x and y are not equal and are interchangeable).
Whether a prime can be expressed in the form of x^2 + Mxy + y^2 where x and y are coprime depends upon the value of M. If M = 2 all integers n have upto (n-1)/2 distinct solutions But, if M = 1 and $$n = p_{a1}^{i1}*p_{a2}^{i2}*..p_{ah}^{ih} where p are primes, believe that there are k and only k distinct solutions where x and y are positive coprime integers (k = the sum of the powers {i1,i2,...ih} or = the number of prime factors but I don't have my work before me and I forgot) if and only if there exists solutions at all. But for M = 1 certain primes such as 3,5,11... and their products can not be expressed in this form at all. For M > 2 if x,y is a solution for n then y and (My -x) is also a solution. Note that the first few solutions for x^2 + Mxy + y^2 = n^2 occur only for n = a prime p for M > 2 since my conjecture is that if n is a product of primes then each of the primes that divide n and the the powers and products of those primes can also be expressed in the form x^2 + Mxy + y^2 for that M. Last edited: Just calculated a interesting result If [tex] n = a^2 + Mab + b^2$$,$$c = b^2-a^2$$ and $$d=2ab -Mb^2$$

then $$n^2 = c^2 -Mcd + d^2$$

also the polarity of any 2 of M, c and d can be reversed since -1 * -1 = 1

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The above formula doesn't seem to work. Let a=b=M=1. Give n=3, c=0, d=1. Then it is not true: 3^2=0-0+1^2.

In the case of a^2+b^2+ab, we can look at this in the form: $$\frac{a^3-b^3}{a-b}.$$ So for a not equal to b, the case for 3, we have the form: $$a^3\equiv{b^3}\bmod{p}$$ where the two integers are not the same. This shows that 3 is a divisor of p-1, and therefore p=3k+1. Since k is even, the form is 6k+1.

This then includes the primes 7, 13, 19, 31...As suggested by Ynaugh?

Whether there is a strictly positive integral form for the square of a prime is another matter. As ramsey2879 shows in his original entry: "Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49."

So here we have the number 7, of the form 3k+1, and 7=2^2+1^2+2x1, and so there is an all positve form for the square.

Also, I see that 13=3^2+1^1+3, and 13^2 = 8^2+7^2+56. Now, 19=3^2+2^2+6; 19^2=16^2+5^2+80.
31=5^2+5+1; 31^2=24^2+11^2+264.

There is, however, a key to this matter: We can decompose $$a^2+b^2+ab = (a-b\omega)(a-b\omega^2)$$, where $$\omega =\frac{-1+\sqrt-3}{2}$$

For example: finding the case of 37 = 4^2+3^2+12. We decompose this into:$$(4-3\omega)(4-3\omega^2)$$ then we square the first factor and simplify using the fact: $$1+\omega+\omega^2=0$$. We then arrive at the first factor for the square is: $$(7-33\omega)$$

The square solution then is: $$37^2=33^2+7^2+33x7$$

The use of this method then makes it clear, anytime we can find the above form for the prime, of the form 6k+1, we can build on this to any higher power as well as to any product of such primes. For example: $$43^3=126^2+197^2+126x197.$$

However, if we are to have an all positive form then a and b must be of the different signs in the decomposition. Three is a problem in itself, $$3=(1-\omega)(1-\omega^2)$$, and there is no good form for 9, trivally, 3^2=3^2+3^3-3^2. Using the above ideas to get the form 3 times 37^2, we arrive at $$-26-63\omega$$, this produces $$3x37^2=26^2+73^2-26x73$$

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Hey Robert Ihnot, I am a bit slow... Can you expand on this?

robert Ihnot said:
There is, however, a key to this matter: We can decompose $$a^2+b^2+ab = (a-b\omega)(a-b\omega^2)$$, where $$\omega =\frac{-1+\sqrt-3}{2}$$

Thanks!

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O.K., actually I thought you were way ahead of me on this!

Omega is the cube root of 1, which can be gotten from Euler's formula, here we have:
cos(120) +isin(120). In an equation such as X^3-1 = 0. The second quadratic term, x,which is the sum of the roots, goes out. That is cos(120)+isin(120) +cos(240)+isin(240) =-1, so adding the third root cos(360)+isin(360) becomes just 0. (This may not be your question, but as a general rule the sum of the nth roots of unity is zero.)

Multiply together$$(a-b\omega^2)a-(a-b\omega^2)b\omega = a^2 +-ab\omega^2-ab\omega+b^2.$$ Now for the terms: $$-ab\omega-ab\omega^2=ab$$ So that we are left with $$a^2+ab+b^2$$ Thus what has happened has been a factoring over the complex numbers.

Another case of this is over the Gaussian integers we have:$$X^2+Y^2=(X+iY)(X-iY)$$

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robert Ihnot said:
The above formula doesn't seem to work. Let a=b=M=1. Give n=3, c=0, d=1. Then it is not true: 3^2=0-0+1^2.

In the case of a^2+b^2+ab, we can look at this in the form: $$\frac{a^3-b^3}{a-b}.$$ So for a not equal to b, the case for 3, we have the form: $$a^3\equiv{b^3}\bmod{p}$$ where the two integers are not the same. This shows that 3 is a divisor of p-1, and therefore p=3k+1. Since k is even, the form is 6k+1.
a and b are coprime and also can not be both equal to 1, Question is 1 coprime to 1? Origionally I did not consider this exception. Thanks for pointing it out.

robert Ihnot said:
O.K., actually I thought you were way ahead of me on this!
LOL I am just trying to understand a picture and I did not mean to hi-jack the thread...

Omega is the cube root of 1, which can be gotten from Euler's formula, here we have:
cos(120) +isin(120). In an equation such as X^3-1 = 0. The second quadratic term, x,which is the sum of the roots, goes out. That is cos(120)+isin(120) +cos(240)+isin(240) =-1, so adding the third root cos(360)+isin(360) becomes just 0. (This may not be your question, but as a general rule the sum of the nth roots of unity is zero.)

Thank you. That helped.

1s are coprime? Well, the important matter is that if p, a prime, =a^2+b^2+ab, it is obvious that a and b could not have a common factor greater than 1. And the case for 3 is that a=b=1.

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robert Ihnot said:
1s are coprime? Well, the important matter is that if p, a prime, =a^2+b^2+ab, it is obvious that a and b could not have a common factor greater than 1. And the case for 3 is that a=b=1.
Yes but c^2 -cd + d^2 <> 9 for c and d as integers. Also, the important matter is that a <> b. How about n is not prime if the greatest commom divisor of c and d <> 1 since the greatest common divisor of c and d divides n If gcd(c,d) = 1 then n^2 = c^2 -Mcd + d^2?

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You can work on that!

robert Ihnot said:
The above formula doesn't seem to work. Let a=b=M=1. Give n=3, c=0, d=1. Then it is not true: 3^2=0-0+1^2.

In the case of a^2+b^2+ab, we can look at this in the form: $$\frac{a^3-b^3}{a-b}.$$ So for a not equal to b, the case for 3, we have the form: $$a^3\equiv{b^3}\bmod{p}$$ where the two integers are not the same. This shows that 3 is a divisor of p-1, and therefore p=3k+1. Since k is even, the form is 6k+1.

This then includes the primes 7, 13, 19, 31...As suggested by Ynaugh?

Whether there is a strictly positive integral form for the square of a prime is another matter. As ramsey2879 shows in his original entry: "Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49."

So here we have the number 7, of the form 3k+1, and 7=2^2+1^2+2x1, and so there is an all positve form for the square.

Also, I see that 13=3^2+1^1+3, and 13^2 = 8^2+7^2+56. Now, 19=3^2+2^2+6; 19^2=16^2+5^2+80.
31=5^2+5+1; 31^2=24^2+11^2+264.

There is, however, a key to this matter: We can decompose $$a^2+b^2+ab = (a-b\omega)(a-b\omega^2)$$, where $$\omega =\frac{-1+\sqrt-3}{2}$$

For example: finding the case of 37 = 4^2+3^2+12. We decompose this into:$$(4-3\omega)(4-3\omega^2)$$ then we square the first factor and simplify using the fact: $$1+\omega+\omega^2=0$$. We then arrive at the first factor for the square is: $$(7-33\omega)$$

The square solution then is: $$37^2=33^2+7^2+33x7$$

The use of this method then makes it clear, anytime we can find the above form for the prime, of the form 6k+1, we can build on this to any higher power as well as to any product of such primes. For example: $$43^3=126^2+197^2+126x197.$$

However, if we are to have an all positive form then a and b must be of the different signs in the decomposition. Three is a problem in itself, $$3=(1-\omega)(1-\omega^2)$$, and there is no good form for 9, trivally, 3^2=3^2+3^3-3^2. Using the above ideas to get the form 3 times 37^2, we arrive at $$-26-63\omega$$, this produces $$3x37^2=26^2+73^2-26x73$$
Now that I had more time to consider this I did make a typo in the formula for d it should have been 2ab +mb^2 so my form reduces to 0^2 -0*3 + 3^2 = 3^2

Note That If you square $$a-b\omega$$ you get $$a^2 - 2ab\omega + b^2\omega^2 = a^2-b^2 - (2ab+b^2)\omega$$ so c = a^2-b^2 and d = (2ab+b^2).

Thus c^2 +cd + d^2 = [-c]^2 -[-c]d + d^2 = n^2 which is my result.

Also this proves that gcd(c,d) divides n since each term on the left of my equation is divisable by the square thereof.

A interesting portion of my conjecture is that if n is composite then each factor of n must be of the form a^2 + Mab + b^2 but as you said in the last post M = 13 is a problem since a = 0 and b = 3 is a trival result, which I think is removed by the requirement that gcd(c,d) must equal 1. However, your showing that $$3x37^2=26^2+73^2-26x73$$ needs to be considered. There $$c = 73^2 - 26^2 = 4653$$ Since 3|4653 then the gcd(c,d) at least equals 3. Thus this does not contradict my conjecture that all n of the general form where gcd(c,d) = 1 can be expressed as a product of primes each of which is of the form where the gcd(c,d) = 1.

PS Do you have any suggestion of how to extend this to the general form where |M|>2?

If my conjecture is true this result can find large primes by keeping a = 1, b = 2 and choosing a large M not divisible by 3 such that the gcd(c,d) = 1.

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ramsey2879 said:
If my conjecture is true this result can find large primes by keeping a = 1, b = 2 and choosing a large M not divisible by 3 such that the gcd(c,d) = 1.
I just realized that if a = 1 and b = 2, gcd(c,d) = gcd(2ab + Mb^2,b^2-a^2) which equals gcd(-3,4+4M) = 1 for M = 0 or 1 mod 3, but 5 + 2M is not prime for all M = 0 or 1 mod 3. I need to consider this further.

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hj
ramsey2879 said:
Now that I had more time to consider this I did make a typo in the formula for d it should have been 2ab +mb^2 so my form reduces to 0^2 -0*3 + 3^2 = 3^2

Note That If you square $$a-b\omega$$ you get $$a^2 - 2ab\omega + b^2\omega^2 = a^2-b^2 - (2ab+b^2)\omega$$ so c = a^2-b^2 and d = (2ab+b^2).

Thus c^2 +cd + d^2 = [-c]^2 -[-c]d + d^2 = n^2 which is my result.

Also this proves that gcd(c,d) divides n since each term on the left of my equation is divisable by the square thereof.

...
If my conjecture is true this result can find large primes by keeping a = 1, b = 2 and choosing a large M not divisible by 3 such that the gcd(c,d) = 1.
I found that if we let $$a = 2$$ and $$b = 1$$ then any odd number greater than 5 can be expressed in the form $$N = 5 + 2M = a^2 + Mab + b^2$$ and so the square is of the form $$N^2 = c^2 + cdM + d^2$$ where $$c = 3$$ and $$d = 4 + M$$. Of course $$M = (N -5)/2$$.

Now If 3|N then 3|d and gcd(c,d) $$\neq$$ 1 but if 3 does not divide N the gcd(c,d) = 1. I considered the relation between c,d and the factors of $$N$$.
THEOREM: If and only if $$R$$ is a factor of N then $$gcd(c+2t,d+t) = gcd(3 + 2t,(N-5)/2 + 4 + t) = R$$ for $$t = (R-3)/2$$
Proof: For $$t equals (R-3)/2$$, $$c + 2t = R$$ and $$d = (N + R)/2and$$gcd(a,n + a) = a[/tex] only if a|n.

thus 35 = 5 + 2*15, c = 3, d = 15+4 = 19

t c+2t d + t gcd(c+2t,d+t)
0 3 19 1
1 5 20 5
2 7 21 7
3 9 22 1
4 11 23 1
5 13 24 1
6 15 25 5
7 17 26 1
8 19 27 1
9 21 28 7
10 23 29 1
..
16 35 35 35

As seen the algorithm starts with the gcd of a numbers on the order of N/2 and 3 and proceeds in an orderly fashion to find the divisors of N. Is this useful or is it too much effort to be useful?

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## 1. What is a conjecture?

A conjecture is a statement or idea that is believed to be true, but has not been proven yet. It is based on observations or evidence, but it still requires further investigation to be confirmed.

## 2. What does "form x^2 + Mxy + y^2" mean?

This is a mathematical expression known as a quadratic form. It consists of three terms, with the variables x and y raised to the second power and a coefficient M in the middle term.

## 3. How is the conjecture related to this quadratic form?

The conjecture suggests that for any integer values of x and y, the expression x^2 + Mxy + y^2 will always result in a perfect square number. In other words, it will always be possible to find two integers that, when substituted into the expression, will give a square number as the result.

## 4. Has this conjecture been proven?

No, this conjecture has not been proven yet. It is still considered an open problem in mathematics and has been a subject of research for many years.

## 5. What are the potential applications of this conjecture?

If proven to be true, this conjecture could have significant implications in number theory and algebra. It could also have practical applications in fields such as cryptography and coding theory.

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