Prove that there are infinitely many integers that are not of the form $n^3+2m^2$ for integers $m,n$. (Hint: Consider the situation modulo 8).

Question

Can I get verification on my proof? Thank you.

Prove that there are infinitely many integers that are not of the form $n^3+2m^2$ for integers $m,n$. (Hint: Consider the situation modulo 8).

We consider the situation modulo $8$: \begin{array}{c|c} n\downarrow,\; m\rightarrow & 0\quad1\quad2\quad3\quad4\quad5\quad6\quad7\quad mod\;8 \\ \hline 0 & 0\quad2\quad0\quad2\quad0\quad2\quad0\quad2\quad\quad\quad\;\; \\ 1 & 1\quad3\quad1\quad3\quad1\quad3\quad1\quad3\quad\quad\quad\;\; \\ 2 & 0\quad2\quad0\quad2\quad0\quad2\quad0\quad2\quad\quad\quad\;\; \\ 3 & 3\quad5\quad3\quad5\quad3\quad5\quad3\quad5\quad\quad\quad\;\; \\ 4 & 0\quad2\quad0\quad2\quad0\quad2\quad0\quad2\quad\quad\quad\;\; \\ 5 & 5\quad7\quad5\quad7\quad5\quad7\quad5\quad7\quad\quad\quad\;\; \\ 6 & 0\quad2\quad0\quad2\quad0\quad2\quad0\quad2\quad\quad\quad\;\; \\ 7 & 7\quad1\quad7\quad1\quad7\quad1\quad7\quad1\quad\quad\quad\;\; \\ \end{array} Then we can see that $4$ is missing from the table. This means that all integers of the form $8k+4$ cannot be $n^3+2m^2$ for integers $m,n$. Thus, it is clear that there are infinitely many integers that are not of the form $n^3+2m^2$.

Answer 1

It's not clear why you write mod $4$ instead of mod $8$ in the last paragraph. I also don’t understand why you write “without getting $n^3+2m^2$”, since the table shows which residues of $n^3+2m^2$ you do get. Other than that, your proof looks fine. Note that $6$ is also missing in the table.

Answer 2

You are correct.

Without a table, you can instead see that if $n^3+2m^2$ is divisible by $4$ but not by $8$

then $n$ must be even

so $n^3$ is divisible by $8$

so $2m^2$ is divisible by $4$ but not by $8$

so $m^2$ is divisible by $2$ but not by $4$

which is impossible because if $m$ is even then $4$ divides $m^2$ while if $m$ is odd then $m^2$ is odd.