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$ x^2+\sqrt{d}x+d = 0 $ ,

Where $d$ is the real number whose integer part is $0$ and it's $n$-th decimal digit is given by (for $n>0$):

$ \text{NthDecimalDigit}(n) = \left \{ \begin{array}{l}0,\: \:\: \text{if $2n+2$ can be expressed as the sum of two primes}\\ 1, \:\: \: \text{otherwise.} \end{array} \right.$

Obviously, the discriminant is -3d and d by definition is non-negative. Therefore, it holds that:

The equation has a real solution (x=0) if and only if d=0 (1)

But obviously, it also holds that:

d = 0 if and only if Goldbach's conjecture is true.

It is commonly stated that "the quadratic formula can solve every quadratic equation" (I provide a proof that this is not the case). Anyway, what do you think is the strict answer to the question of the title (some scientists consider equivalence (1) as a "solution")?

George Bougioukas
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1 Answers1

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Let $$x^2 + px + q = 0$$ a quadratic equation then the discrimant is given by $\sqrt{\frac{p^2}{4} - q}$ so in your case it's $\sqrt{\frac{d}{4} - d}$

Because $d>0$ it holds $\frac{d}{4} - d < 0$ so your equation has no solution.

Gono
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  • Can you prove that d>0? – George Bougioukas Dec 06 '16 at 20:23
  • It holds $$d = \sum_{n=1}^\infty \frac{a_n}{10^n}$$ with $$a_n = \begin{cases} 0 & \text{if $2n+2$ can be expressed as the sum of two primes}\ 1 & \text{otherwise}\end{cases}$$

    If you mean by "can be expressed as the sum of two primes" that two different primes are needed $a_1 = 1$ so $d \ge 0.1$.

    If not your question depends on Goldbachs Conjecture… if he's right $d=0$ and your equation has one exact solution: $x=0$. If he's not right $d>0$.

     – Gono Dec 06 '16 at 20:26
  • Yes, it depends on the Goldbach's conjecture. So, you claim the solution of the equation is: "if he's right d=0 and your equation has one exact solution: x=0. If he's not right d>0" – George Bougioukas Dec 06 '16 at 20:29
  • Edited my comment, was sent to early… but it doesn't matter: if $d=0$ you have one exact solution, if $d\not=0$ then $d>0$ and you have no solution… – Gono Dec 06 '16 at 20:31
  • You mean these material implications are solutions?? – George Bougioukas Dec 06 '16 at 20:37
  • For your simple reformulation of Goldbachs conjecture… yeah :-) – Gono Dec 06 '16 at 21:16
  • So, you've just solved the Goldbach's conjecture :) – George Bougioukas Dec 06 '16 at 21:22