How to solve this: Is $(1782^{12}+1841^{12})^{1/12}$ an integer? Should I reexpress $1841$ in terms of $1782$, as such: $1782 = x$, and $1841 = x+59$? I can't make any progress. Please help.
Asked
Active
Viewed 46 times
0
-
No! It can not be integer. We know this at least 300 years. – Michael Rozenberg Sep 24 '17 at 20:49
-
Please explain. I have not descended from prehistoric times. – DarkRunner Sep 24 '17 at 20:52
-
Prove that the equation $x^3+y^3=z^3$ has no natural solutions. – Michael Rozenberg Sep 24 '17 at 20:53
-
Fermat proved that no fourth power can be written as the sum of two fourth powers. If your number was an integer $m$, then $(1782^3)^4+(1841^3)^4=(m^3)^4$. – José Carlos Santos Sep 24 '17 at 20:57
2 Answers
3
According to Fermat's last theorem, there are no solutions for $$c^n = a^n + b^n$$ when a, b, and c are integers and n is an integer greater than 2. If $$\sqrt[12]{(1782^{12}+1841^{12})} = c$$ for some integer c, that would imply that $$1782^{12}+1841^{12} = c^{12}$$ which we know is impossible.
2.71828-asy
- 417
0
Fermat's Last Theorem works works, but so do more elementary approaches.
We have $1782^{12}=(1782^3)^4$ which by Fermat's Little Theorem is $\equiv 1 \bmod 5$. Likewise for $1241^{12}$. Then a proposed integer $c$ such that $c^{12}=1782^{12}+1241^{12}$ would imply that $c^{12}=(c^3)^4\equiv 2 \bmod 5$, which is contradicted by Fermat's Little Theorem.
Oscar Lanzi
- 39,403