Numar care nu este patrat perfect

[mihai miculita]

Aratati ca numarul $32^{20}+23^{22}$, nu este un patrat perfect!

[Stefan Dominte]

Considerand expresia modulo 3, obtinem ca aceasta este M3 +2, prin urmare nu poate fi patrat perfect.

[Stefan Tudose]

De asemenea, $32^{20}+23^{22}$ este congruent cu 2 modulo 11.

Dar $\left(\dfrac{2}{11}\right)=(-1)^{\frac{11^2-1}{8}}=(-1)$, deci $32^{20}+23^{22}$ nu este patrat perfect.

[mihai miculita]

SOLUTIE PRIN REDUCERE LA ABSURD:
Presupunem ca exista $n\in\mathbb{N}^*$, astfel incat sa avem:
$32^{20}+23^{22}=n^2\Rightarrow$
$\Rightarrow23^{22}=n^2-32^{20}=n^2-(2^5)^{20}=n^2-2^{100}=$
$=(n-2^{50})(n+2^{50})\Rightarrow\left\{\begin{array}{c}n+2^{50}=23^k\\ n-2^{50}=23^{22-k};\,k>11\end{array}\right\}\Rightarrow$$2^{51}=(n+2^{50})-(n-2^{50})=23^k-23^{22-k}\Rightarrow$
$\Rightarrow k=22\Rightarrow 2^{51}=23^{22}-1\Rightarrow$
$\Rightarrow 1=23^{22}-2^{51}>(2^4)^{22}-2^{51}=2^{88}-2^{51}=$$2^{51}.(2^{37}-1)>2^{51}\Rightarrow \boxed{1>2^{51};\,\mbox{Fals!}}\Rightarrow$
$\Rightarrow \mbox{Presupunerea facuta este FALSA!}$
SOLUTIE DIRECTA:
1). $32^{10}=(32^2)^{10}=1024^{10};$
Insa: $1024\in\{25k-1\}\Rightarrow 1024^2\in\{25k+1\}\Rightarrow$$\boxed{32^{20}=1024{10}\in\{25k+1\}}.\,\,(1)$
2). $23^2=529\in\{25k+4\}\Rightarrow$
$\Rightarrow 529^{2}\in\{25k+16\}\Rightarrow 529^{4}\in\{25k+6\}\Rightarrow 529^{8}\in\{25k+11\};$
$\Rightarrow 529^{10}\in\{25k+1\}\Rightarrow \boxed{23^{22}=529^{11}\in\{25k+4\}}.\,\,(2)$
Din (1) si (2)$\Rightarrow 32^{20}+23^{22}\in\{25k+5\}\Rightarrow$
$\Rightarrow (32^{20}+23^{22}){\vdots}5$ si $(32^{20}+23^{22})\not{\vdots}25\Rightarrow$$\boxed{32^{20}+23^{22}\, \mbox{ nu este patrat perfect!}}$