Consider the Diophantine equation:
$10^n-a^3-b^3=c^2$, for $a$, $b$, $c$, and $n$ positive.
Has this equation infinitely many solutions?
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3$\begingroup$ I think you really want to provide a motivation for the question and show your own efforts in solving this problem --- to improve the reception here at MO. $\endgroup$– Carlo BeenakkerCommented Mar 21, 2020 at 21:32
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1$\begingroup$ Feels like an olympiad problem. $\endgroup$– hookahCommented Mar 21, 2020 at 22:12
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$\begingroup$ Variation now posted as mathoverflow.net/questions/355415/… $\endgroup$– Gerry MyersonCommented Mar 22, 2020 at 6:02
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1 Answer
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Yes, it has.
Note that using (the truly genius observation!) $10^n = 8\cdot 10^{n-1}+10^{n-1}+10^{n-1}$; it suffices to choose $n\equiv 1\pmod{3}$, and $n\equiv 1\pmod{2}$. That is, take $n=6\ell+1$, $\ell\in\mathbb{N}$. Then, $$ 10^n=10^{6\ell+1} = (2\cdot 10^{2\ell})^3 + (10^{2\ell})^3 + (10^{3\ell})^2. $$ Thus, $$ (a,b,c,n)=(2\cdot 10^{2\ell},10^{2\ell},10^{3\ell},6\ell+1),\quad \ell\in\mathbb{N} $$ is a parametric family along which you get infinitely many solutions.
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$\begingroup$ are there solutions with a, b c >1 and not multiple of 10? $\endgroup$ Commented Mar 21, 2020 at 22:27