INICIANTE:
Por absurdo, suponha que exista. Em particular,
não é múltiplo de
. Elevando a congruência à
-ésima potência e aplicando o Teorema de Fermat, obtemos:
![a](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_0cc175b9c0f1b6a831c399e269772661.gif?w=640&ssl=1)
![p](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_83878c91171338902e0fe0fb97a8c47a.gif?w=640&ssl=1)
![\dfrac{p-1}{2}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_feb5141edbba7583a98e006a49012a31.gif?w=640&ssl=1)
![a^{p-1} \equiv (-1)^{\dfrac{p-1}{2}} \pmod{p}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_1c54d498aeb208c0f4ec905a3ad434a5.gif?w=640&ssl=1)
![1 \equiv -1 \pmod{p}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_53b5ab063b67786699022b22b6fcd06b.gif?w=640&ssl=1)
pois
é ímpar. Isso nos dá a contradição requerida.
![\dfrac{p-1}{2}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_feb5141edbba7583a98e006a49012a31.gif?w=640&ssl=1)
INTERMEDIÁRIO:
Da relação do problema, temos:
![a^3=6(a+1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_954d8b4d8680cafd227e3554463bb907.gif?w=640&ssl=1)
![\Rightarrow a^2=\dfrac{6(a+1)}{a}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_bfeb04e473189dcd0141ed9e817f3753.gif?w=640&ssl=1)
pois
é positivo. Por absurdo, se a equação do segundo grau requerida tem solução real, então seu discriminante é não-negativo:
![a](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_0cc175b9c0f1b6a831c399e269772661.gif?w=640&ssl=1)
![a^2 - 4(a^2-6) \ge 0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_69f472372ce38ca139bcd5779a413636.gif?w=640&ssl=1)
![-3a^2+24 \ge 0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_043cdb638242591603ff61e29dcc5c6f.gif?w=640&ssl=1)
![8 \ge a^2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_f2ff43d560bc3b68427acb9b2c2052ca.gif?w=640&ssl=1)
o que, por (1) equivale a:
![8 \ge \dfrac{6(a+1)}{a}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_df96bea8345a15290b08db110c009035.gif?w=640&ssl=1)
![a \ge 3](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_634a89ead26ff74721b90c84d8f40f4b.gif?w=640&ssl=1)
![a^2 \ge 9](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_d0b7a3974065d4432f9d899466d22fa3.gif?w=640&ssl=1)
contradizendo
.
![8 \ge a^2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_abd6e7d9dc3300323f55985a05badbea.gif?w=640&ssl=1)
AVANÇADO:
Seja
, digamos
![d=mdc(a,b)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_ee527303b05b2eb9726a5c994ea4bae1.gif?w=640&ssl=1)
![a=d\cdot a_0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_5de1aec9053063b9ac86b5aae5457e66.gif?w=640&ssl=1)
![b=d\cdot b_0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_5dd948cc0acb5b4db3e2974ce22a5257.gif?w=640&ssl=1)
onde o
. Então:
![mdc(a_0,b_0)=1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_3e7c121ab2bb98d9d5688c46ef3fe177.gif?w=640&ssl=1)
![2^{a}-1=2^{d\cdot a_0}-1=(2^{d}-1)\cdot(2^{d(a_0 -1)}+2^{d(a_0-2)}+...+1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_34118fb64d452fd139cf51bf5f9ad376.gif?w=640&ssl=1)
é múltiplo de
. Aplicando a mesma fatoração para
, segue que
![2^d-1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_cffca9b16517ae8badb86c4a0b233f14.gif?w=640&ssl=1)
![2^{b}-1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_3805ad60805eef3f3d7fe81d024ea0a0.gif?w=640&ssl=1)
![2^{d}-1 \div mdc(2^{a}-1,2^{b}-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_fe01acf54fecaf33fa1b4e81c4b6c1db.gif?w=640&ssl=1)
Para a divisibilidade inversa, sejam
inteiros positivos tais que
![x,y](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_f10bc3c94b77e1d6b9f98106daf335c1.gif?w=640&ssl=1)
![d=ax-by](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_c3efb273aff024b64219593dd61f8b87.gif?w=640&ssl=1)
Então
são múltiplos de
e daí
![2^{ax}-1, 2^{by}-1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_5dbfb49f11b9a60098edcce7add79ea2.gif?w=640&ssl=1)
![mdc(2^{a}-1,2^{b}-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_6133eb1f3846281a8b5a021959ef3cf5.gif?w=640&ssl=1)
![mdc(2^{a}-1,2^{b}-1) \div (2^{ax}-1)-(2^{by}-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_43866b33dc84eab87952196ae58e39e3.gif?w=640&ssl=1)
![mdc(2^{a}-1,2^{b}-1) \div 2^{d+by}-2^{by}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_8d3fb44475a15f1bca549d0a905aed65.gif?w=640&ssl=1)
![mdc(2^{a}-1,2^{b}-1) \div 2^{by}(2^{d}-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_f049e58de483ecbc669359d53b758e88.gif?w=640&ssl=1)
![mdc(2^{a}-1,2^{b}-1) \div 2^{d}-1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_330fc1ab55fd65ef34323f67063389ac.gif?w=640&ssl=1)
pois
é ímpar.
![mdc(2^{a}-1,2^{b}-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_6133eb1f3846281a8b5a021959ef3cf5.gif?w=640&ssl=1)