Let $t>1$ be an odd integer. The following system is easy to verify. \begin{align*} \left( \frac{\left(9t^4 \right)^3-1}{2} \right)^2+0 &= \left( \frac{\left(9t^4-3t \right)^3-\left(9t^3-1 \right)^3}{2} \right)^2 + \left(\left(9t^3-1 \right) \left(9t^4-3t \right) \right)^3, \\ \left( \frac{\left(9t^4 \right)^3-1}{2} \right)^2 + 1 &= \left( \frac{\left(9t^4 \right)^3-1}{2} \right)^2 + 1^3, \\ \left( \frac{\left(9t^4 \right)^3-1}{2} \right)^2 + 2 &= \left( \frac{\left(9t^4 \right)^3-1}{2}-1 \right)^2 + \left(9t^4 \right)^3. \end{align*}Done. $\hfill \square$ Now that seems kind of contrived. Let me write another solution that may be more motivated than the above. Check my blog to see how I arrived at that solution itself. Lemma [from here]. For all real $t$ we have \[ \left(9t^3-1 \right)^3+\left(9t^4-3t \right)^3 = \left(9t^4 \right)^3-1. \]Proof. Straightforward. The above is a parametric solution. Just multiply it out. $\hfill \square$ We want to find positive integers $x$ such that $x^2,x^2+1,x^2+2$ are all heinersch. Of course $x^2+1^3$ is heinersch. So we are looking for $x$ such that there exist $a,b,c,d \in \mathbb{Z}_{>0}$ with \[ \begin{cases} \begin{aligned} x^2+2 &= a^2+b^3, \qquad \qquad(1)\\ x^2 &= c^2+d^3. \qquad \qquad(2) \end{aligned} \end{cases} \]For equation (1) take $b>1$ odd, $x = \tfrac{b^3-1}{2}$ and $a=x-1$. It is easy to verify that this always solves (1). For equation (2) we set $x = \tfrac{b^3-1}{2}$ with $b>1$ odd and factorize to \[ \left( \frac{b^3-1}{2} \right)^2-c^2 = d^3 \iff \left( \frac{b^3-1}{2}-c \right) \left(\frac{b^3-1}{2}+c \right) = d^3. \]It suffices to force both factors on the left to cubes. I.e. we want to find $b,c$ such that there exist $m,n \in \mathbb{Z}_{>0}$ with \[ \begin{cases} \begin{aligned} m^3 &= \frac{b^3-1}{2}-c, \\ n^3 &= \frac{b^3-1}{2}+c. \end{aligned} \end{cases} \]Here it suffices to solve the equivalent system \[ \begin{cases} \begin{aligned} m^3+n^3 &= b^3-1, \qquad \qquad \text{ } (\text{i}) \\ n^3-m^3 &= 2c. \qquad \qquad \qquad (\text{ii}) \end{aligned} \end{cases} \]For equation (i) take $b = 9t^4, m = 9t^3-1, n=9t^4-3t$ with odd $t>1$ and use the Lemma. For equation (ii) set $c = \tfrac{n^3-m^3}{2}$. It's easy to see that $c$ is a positive integer. Hence our construction is complete and we are done! $\hfill \square$

\begin{align*} (2b^2+1)^2-1&=(2b^2-2b)^2+(2b)^3 \\ (a^2-1)^2&=(a^2-4a+3)^2+(2a-2)^3 \\ n^2+1&=n^2+1^3 \end{align*}Now, we just have to show that $2b^2+1=a^2-1$ has infinitely many solutions, which is easy (because it is a Pell-type equation).

0.And thus I started exploring unfamiliar fields like matrices or how to do addition. 1. But that was no reason to give up. No. Persistence is the way to go. Did Andrew Wiles give up when a hole was found in his long-awaited proof that he’s worked on for 7 straight years? Of course not. Neither did I.

0.And thus I started exploring unfamiliar fields like matrices or how to do addition. 1. But that was no reason to give up. No. Persistence is the way to go. Did Andrew Wiles give up when a hole was found in his long-awaited proof that he’s worked on for 7 straight years? Of course not. Neither did I.