Can every sufficiently large integer be written in the form $a^2 + b^3 + c^6 $ for some nonnegative integers $a,b$ and $c$?

8$\begingroup$ Please do not delete the content of the question. $\endgroup$ – Gjergji Zaimi Dec 3 '12 at 21:26
This is question 3 on page 146 of the second edition of The HardyLittlewood Method by R. C. Vaughan, with $b \geq 0.$
_{(Image added by J.O'Rourke)}
The answer is No, by a simple volume argument. It is not even necessary to know the exact constant, just that the number of lattice points with $x,y,z \geq 0 $ and $$ x^2 + y^3 + z^6 \leq N$$ is less than $CN,$ with a constant $0 < C < 1.$
I'm in question 5 on the same page.
Image hosting did not work. Please see: VAUGHAN PDF
Oh, well. The relevant calculation is the sum of the reciprocals of the exponents, in case the polynomial is the sum of distinct monomials in the different variables. You might think that $$ x^2 + y^2 + z^9$$ ought to represent all large numbers with $z \geq 0.$ There are no local obstructions. But this is not the case. We can think of the exponent in this volume calcultaion as $10/9.$
It is reasonable to ask, how large an exponent reciprocal sum can we get and still fail to represent large integers? The best I have is $4/3,$ and in the simplest form we require coefficients, as in $$ x^2 + 27 y^2 + 7 z^3. $$ This is a version of the $ 4 x^2 + 2 x y + 7 y^2 + z^3, $ which is more natural but looks less diagonal.

$\begingroup$ @Joseph, thanks. I bought a scanner for home, it is a nice flatbed that does one page at a time. Canon Canoscan LiDE 210. Of course I had to update Ubuntu to get it to run, which disabled my printer. It's always something. So I have been able to post stuff on MSE for months, sometimes line drawings i do, maybe book pages. I tried the same thing here. i thought I did it properly, but no image appeared. $\endgroup$ – Will Jagy Dec 3 '12 at 2:53

$\begingroup$ If I remember right there's a known example of a quadratic form $ax^2+by^2+cz^2$ (so "exponent reciprocal sum" $3/2$) and an arithmetic progression in which every positive term is represented except the squares. (Restriction to an arithmetic progression is the best one can hope for, because it's impossible for a ternary quadratic form over $\bf Z$ to be unobstructed for all positive integers.) $\endgroup$ – Noam D. Elkies Dec 3 '12 at 3:35

1$\begingroup$ @Noam, Yes, of course. The prettiest example would be $3 x^2 + 4 y^2 + 9 z^2,$ which does not represent any $w^2$ where all prime factors of $w$ are $\equiv 1 \pmod 3.$ That goes back to Jones and Pall (1939). The proof is elementary, although now there is an entire industry dealing with the spinor genus and spinor exceptional integers, after Eichler. I guess I ruled those out automatically. A (probably) complete list is at math.rwthaachen.de/~Gabriele.Nebe/LATTICES/index.html#Jagy if you click on the word Table. $\endgroup$ – Will Jagy Dec 3 '12 at 3:51

$\begingroup$ Just to fill things in, for (positive) prime $q \equiv 2 \pmod 3$ there is an integral representation $ 3 x^2 + 4 y^2 + 9 z^2 = q^2.$ With $9$ itself, that settles the representation question for all squares. $\endgroup$ – Will Jagy Dec 3 '12 at 7:00

2$\begingroup$ With prime $q \equiv 2 \pmod 3,$ we may take $q = r^2 + s^2 + 3 t^2$ and, by choosing $\pm s,$ then demand that $ r \equiv s \pmod 3.$ Then we have $$ q^2 = 3 (2 r t)^2 + 4 (rs)^2 + 9 \left( t^2 + \frac{s^2  r^2}{3} \right)^2. $$ $\endgroup$ – Will Jagy Dec 3 '12 at 19:43
The number of positive integer triples $(a,b,c)$ such that $a^2+b^3+c^6\le x$ can be estimated from above by
$$x\int_0^1 \int_0^{(1c^6)^{1/3}}\int_0^{(1b^3c^6)^{1/2}} 1 \, da\, db\, dc\, ,$$ that is $\theta x$ with $\theta <1$, so that a portion of integers with positive density is not representable.

3$\begingroup$ The OP did not say that $b$ was positive... $\endgroup$ – Igor Rivin Dec 2 '12 at 20:28

3$\begingroup$ Are you claiming that $a \ge n^{1/2}$, etc? Why? $\endgroup$ – Felipe Voloch Dec 2 '12 at 20:33

2$\begingroup$ Are you saying that a positive proportion of triples in $[0,x^{1/2}] \times [0,x^{1/3}] \times [0,x^{1/6}]$ have $a^2+b^3+c^6>x$? I guess that's clear. $\endgroup$ – Will Sawin Dec 2 '12 at 21:22

2$\begingroup$ The constant is $$ \Gamma \left( \frac{3}{2} \right) \Gamma \left( \frac{4}{3} \right) \Gamma \left( \frac{7}{6} \right). $$ The technique for evaluating the integral is due to Dirichlet and is in the book by Whitakker and Watson. $\endgroup$ – Will Jagy Dec 2 '12 at 21:45

4$\begingroup$ @Pietro Majer: The numbers which are not a sum of three squares are of the form $4^a(8b+7)$, hence their density is $\frac{1}{8}\sum_a 4^{a}=\frac{1}{8}\frac{4}{3}=\frac{1}{6}$. $\endgroup$ – GH from MO Dec 2 '12 at 23:44