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Recently, I stumbled upon an interesting statement regarding Quadratic Forms. It is quite well-known and, as I will describe briefly, equivalent to Goldbach's conjecture.

Let $p,q$ be odd primes and consider $Q(x,y) = x^2 - y^2$. A straightforward argument shows that there are only two non-negative integer solutions to $Q(x,y) = pq$. Namely, when $q \leq p$,

$\displaystyle (x,y) = \left(\frac{p+q}{2},\frac{p-q}{2}\right)$ and $\displaystyle (x,y) = \left(\frac{pq+1}{2},\frac{pq-1}{2}\right)$.

Now, suppose we go in the other direction. Given an integer $x = n \geq 2$, can we always find some integer $0 \leq y \leq n - 2$ and primes $p,q$ such that $Q(n,y) = pq$? Clearly, if this was the case, then $n^2 - y^2 = pq$ and it follows that $n = p + y$ and $n = q - y$. Hence, $2n = p + q$.

At this point, I would like to point out that I understand that this is no easier to resolve. However, it let me think that there might be an even deeper connection (which may or may not be true) between Quadratic Forms and Goldbach's conjecture.

After some thought, I began looking at Euler's Idoneal numbers. These are positive integers, $D$, such that if an integer is representable as $x^2 \pm Dy^2$ in only one way with $x^2$ coprime to $Dy^2$, then it is guaranteed to be a prime, a power of a prime or twice one of these. Obviously, $D = 1$ is somewhat related to what I said above, but this time the focus is on the possibility of taking prime values.

Next, I thought to consider the quadratic form

$Q(a,b,c,d) = a^2 + D_1 b^2 + c^2 + D_2 d^2$

and integral representations for even integers. Trivially, every integer is represented by this in the case $D_1 = D_2 = 1$, according to Lagrange, but this is of no help. What I wondered is whether I could choose $D_1$ and $D_2$, idoneal numbers, so that there was only one positive integer solution to

$Q(a,b,c,d) = n$

which would imply that $a^2 + D_1 b^2$ and $c^2 + D_2 d^2$ are representable in only one way and, hence, (possibly) prime numbers. Of course, I mean one way up to $Q(a,b,c,d) = Q(c,b,a,d)$ (and possibly more cases depending on $D_1 = D_2$) and there is no guarantee that $(a^2, D_1 b^2) = (c^2, D_2 d^2) = 1$ must be the case.

In some cases, this is quite easy. For example,

$54 = a^2 + 12b^2 + c^2 + 40d^2$

has the only solution $(a,b,c,d) = (1, 1, 1, 1)$ in positive integers. It is easily seen that this example is useless in general: the coefficient $40$ is quite large compared to $54$ together with the fact that $6, 14, 42$ and $54$ are not sums of two squares; this forces no other solutions. I don't know whether there can be many numbers for which the above works.

I realise that this approach is very weak. In fact, I read a few pages off notes by Pete Clark which suggested that the number of solutions to an equation $q(x_1, x_2, ..., x_n) = N$, for a positive-definite integral quadratic form $q$, is asymptotically $N^{n/2-1}$. I interpret this as saying that, in general, the above approach fails: there is only a certain number of those quadratic forms, since there are only 65 idoneal numbers known.

My questions are the following.

Is it possible that Euler's idoneal numbers, together with some more results (say Gauss' three-square theorem), could imply Goldbach's conjecture for many (some class of) numbers?

Is there a criterion, or something of that nature, which allows one to identify integers that are (essentially) uniquely represented by a given quadratic form?

I will appreciate any references on any related topics. Thank you!


I apologise if something in my post doesn't make sense. I am yet to begin my undergraduate degree in Maths next academic year, but I plan to complete an essay project on a similar topic this summer.

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1 Answer

up vote 4 down vote accepted

There should be only finitely many such $n$. The reason is that there should be too many pairs of primes $(p_i,q_i)$ such that $p_i=a_i^2+b_i^2 D_1$ and $q_i=c_i ^2+d_i ^2 D_2$, and $n=p_i+q_i$ and if we have just two different such representations then we do not get the uniqueness that you want.

If a prime $ p= a ^ 2 + b ^ 2 D $ (where $D$ is such an idoneal number (Cox calls these numbers Eulers convenient numbers, See Cox, Primes of the form $p=x^2+n y^2$, Corollary 2.27 and page 59-60, which I believe might be one of the best references for these questions)), then in particular any prime $q$ such that $q \equiv p \pmod {4D}$ should also be representable in the same form, i.e. $ q = \alpha ^ 2 + \beta^2 D $.

Then my assertion follows from a variant of Goldbach conjecture for arithmetic progressions, such that given $a \pmod m$ and $b \pmod m$ , $(m,a)=(m,b)=1$, then there should exists an $n_0$ such that for each even integer $n \geq n_0$ where $n \equiv a+b \pmod m$ there should exist at least two distinct representations $ n = p _ 1 + q _ 1 = p _2 + q_2 $ where $p_i,q_i$ are prime numbers, $ p_i \equiv a \pmod m $ and $ q_i \equiv b \pmod m $. Although not proven since it is more general than Goldbach's conjecture, it should hold (and also we get an asymptotic for the expected number of representations) by probabalistic/heuristic reasoning (Cramer model...)).

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Thank you for your answer! My feeling was that there would not be many of those $n$, for which this method could work, and now I have a good way of reasoning about it. I will attempt to write a computer program to see what comes out; maybe it will find most of them. By any chance, do you know of a public lecture (available online) or an expository article about the variant of Goldbach's conjecture for arithmetic progressions? I didn't succeed in finding much about it. Thanks! –  M.G. Jun 2 '13 at 18:54
    
I do not know about it. In fact I am not sure if anyone formulated it before, but is a natural conjecture to make so it is likely out there somewhere. Note that this is not a proof (just a reason why there should only be finitely many n). To prove that there are just finitely many cases might be possible with some other method (the reasoning above will not help with a proof I am afraid). –  Johan Andersson Jun 2 '13 at 19:53
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