Is it correct that if an natural number $10^n+89$ is a perfect square then $n=3$?
The answer is clear if $n$ is an even number. For odd $n$ I can prove that $10^n+89$ can be a perfect square only if $n=22m+3$...
2026-03-26 09:40:15.1774518015
On
Can $10^n+89$ ever be a Perfect Square for $n>3$?
371 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
2
There are 2 best solutions below
2
On
This is a semi-proof that there are only finitely many solutions. I use the abc-conjecture here, which has so far not been confirmed to be proven (although Shinichi Mochizuki claimes to have a proof, hence my "semi-proof").
The abc-conjecture.
Given $a,b,c>0$ with $a+b=c$ and $\gcd(a,b,c)=1$, for every $\epsilon>0$ there exists a $K_\epsilon$ such that $$c<K_\epsilon\operatorname{rad}(abc)^{1+\epsilon}$$ where $\operatorname{rad}(x)$ denotes the product of the distinct prime factors of $x$.
So given that $10^n+89=s^2$, we know that for every $\epsilon>0$ there exists a $K_\epsilon$ such that
$$s^2<K_\epsilon\operatorname{rad}(10^n\cdot 89\cdot s^2)^{1+\epsilon}$$
note that $\operatorname{rad}(abc)=\operatorname{rad}(a)\operatorname{rad}(b)\operatorname{rad}(c)$ (for pairwise coprime $a,b,c$, and notice that $10^n$, $89$ and $s^2$ are pairwise coprime). So \begin{align} s^2&<K_\epsilon\operatorname{rad}(10^n\cdot 89\cdot s^2)\\ &=K_\epsilon\left(\operatorname{rad}(10^n)\operatorname{rad}(89)\operatorname{rad}(s^2)\right)^{1+\epsilon}\\ &=K_\epsilon\cdot\left(10\cdot 89\cdot \operatorname{rad}(s)\right)^{1+\epsilon}\\ &=K_\epsilon\cdot890^{1+\epsilon}\operatorname{rad}(s)^{1+\epsilon}\\ &\leq K_\epsilon\cdot890^{1+\epsilon}s^{1+\epsilon} \end{align} which means that
$$s^2<K_\epsilon\cdot890^{1+\epsilon}s^{1+\epsilon}$$
and dividing by $s^{1+\epsilon}>0$ gives
$$s^{1-\epsilon}<K_\epsilon\cdot890^{1+\epsilon}$$
for a fixed $\epsilon>0$, $K_\epsilon$ is fixed, and so is $890^{1+\epsilon}$, so there are at most finitely many solutions for $s$ (since $s$ is positive and bounded above by $\sqrt[1-\epsilon]{K_\epsilon890^{1+\epsilon}}$). Unfortunately, we don't know what $K_\epsilon$ is (except the lower bound $K_\epsilon>\frac{33}{890}29370^{-\epsilon}$ which we can find using the only solution we have), so unfortunately we still can't rule out those finitely many posibilities to get only $s=33$ (that's the solution $n=3$).
There was a good point made by mathlove, the restriction of the base to a fixed number $(10)$ allows calculating three Mordell curves. The bad news is that neither Magma nor Sage is willing to do the final one. I will post the first two. Looking again, it appears the OP also was able to find solutions for the first two curves, but not the third. Looking through articles on computation written after 2005, say, it appears that most curves $y^2 = x^3 + k$ with $|k| \leq 100000$ have been done, but maybe not all, and in any case are not in any convenient websites. It would appear that $|k| > 100000$ makes it likely that nobody has finished the job.
Method: I had not written it down, it just looked correct. However, if $$ y^2 = 10^{6k+3} + 89, $$ take $x = 10^{2k+1}$ to get $$ y^2 = x^3 + 89. $$ If $$ u^2 = 10^{6k+1} + 89, $$ $$ 100u^2 = 10^{6k+3} + 8900, $$ take $y = 10 u, \; \;x = 10^{2k+1}$ to get $$ y^2 = x^3 + 8900. $$ If $$ v^2 = 10^{6k+5} + 89, $$ $$ 10000v^2 = 10^{6k+9} + 890000, $$ take $y = 100 v, \; \;x = 10^{2k+3}$ to get $$ y^2 = x^3 + 890000. $$
Meanwhile, note that there is an excellent selection of literature on Ramanujan-Nagell and Lebesgue-Nagell. https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation The bad news is that the vast majority of explicit solutions is for $x^2 + C = y^n$ where $C>0.$ For those, many relevant articles can be downloaded for free.
TWO OUT OF THREE MORDELL CURVES
Let me throw in