INICIANTE:
![a^2+b^2+(a+b)^2 = c^2+d^2+(c+d)^2 \Rightarrow 2a^2+2b^2+2ab=2c^2+2d^2+2cd \Rightarrow a^2+b^2+ab=c^2+d^2+cd](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_ac9d8b4620f70b64acdff245afeb78ee.gif?w=640&ssl=1)
Eleve ao quadrado ambos os lados:
![(a^2+b^2+ab )^2 = ( c^2+d^2+cd )^2 \Rightarrow a^4 + b^4 + a^2b^2 + 2( a^3b + ab^3 + a^2b^2 ) = c^4 + d^4 + c^2d^2 + 2( cd^3 + c^3d + c^2d^2 )](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_1a2a0393201057d650d973240165cacd.gif?w=640&ssl=1)
Multiplique ambos lados por
:
![2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_c81e728d9d4c2f636f067f89cc14862c.gif?w=640&ssl=1)
![2a^4 + 2b^4 + 2a^2b^2 + 4( a^3b + ab^3 + a^2b^2 ) = 2c^4 + 2d^4 + 2c^2d^2 + 4( cd^3 + c^3d + c^2d^2 )](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_8f4049a2af28472656b1c80ee8ad1878.gif?w=640&ssl=1)
![a^4 + b^4 + ( a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 ) = c^4 + d^4 + ( c^4 + 4c^3d + 6c^2d^2 + 4cd^3 + d^4 )](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_18c2acd8126c29a1fddba026c292e28c.gif?w=640&ssl=1)
![a^4+b^4+(a+b)^4=c^4+d^4+(c+<wbr data-recalc-dims=](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_5912301ddd49deb70933cd8433d6d771.gif?w=640&ssl=1)
INTERMEDIÁRIO:
Note que o lado esquerdo de todas equações é positivo (todo quadrado é positivo), assim, as soluções do sistema devem ser todas positivas. Podemos afirmar sem perda de generalidade que:
![a \ge b \ge c \ge d \ge 0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_7438bd6de17c0d2fc9266b1d95f15ddc.gif?w=640&ssl=1)
Defina
e note que:
![S = a+b+c+d](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_cd8b18cc023333dfc3639b73d5f546df.gif?w=640&ssl=1)
![(S-a)^{2016} = 3a](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_cd6acff3a83339b3eaddeb4e412da986.gif?w=640&ssl=1)
![(S-b)^{2016} = 3b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_9f02b528c3d8d1ca087ba360e441f8b1.gif?w=640&ssl=1)
Como
:
![a \ge b \ge 0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_bebe92d1b9a489d448a440fcb7adf8e2.gif?w=640&ssl=1)
![a \ge b \Rightarrow 3a \ge 3b \Rightarrow (S-a)^{2016} \ge (S-b)^2016](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_d9d549f43d47a0cd41a62b74e0d0dfe4.gif?w=640&ssl=1)
Como
são positivos (a soma de todos é maior que cada um):
![a,b,c,d](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_6b2c3772e956bd7d191e4e9417c49700.gif?w=640&ssl=1)
![(S-a)^{2016} \ge (S-b)^2016 \Rightarrow S-a \ge S-b \Rightarrow b \ge a](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_52b7162b3350f22245118dda66b7bd46.gif?w=640&ssl=1)
Mas
, logo:
.
![a \ge b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_a208c01e821b96c0c957f485907a65b4.gif?w=640&ssl=1)
![a=b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_7acaac15494e6820b1ed6d8b539af089.gif?w=640&ssl=1)
Tomando as desigualdades
e
concluímos analogamente que:
![b \ge c](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_13dc4929cfac4cc7a41d629180e9e8ad.gif?w=640&ssl=1)
![c \ge d](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_c4ebb7d07d339edc5cc43b5054d52526.gif?w=640&ssl=1)
![b=c](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_627c4b2ef5e9574e7c6e35ddc2537965.gif?w=640&ssl=1)
![c=d](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_39527c5de24326c584201a3d28e55a68.gif?w=640&ssl=1)
![\Rightarrow a=b=c=d](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_2f682e53e9e19bc089d7405bf86cbd67.gif?w=640&ssl=1)
Substituindo na primeira equação do sistema:
![(3a)^{2016} = 3a](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_0b89348f4ae7142900db614e9e58548b.gif?w=640&ssl=1)
![(3a)((3a)^{2015} - 1) = 0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_1256b596bb99c8d8fe0b8b16b2f54e88.gif?w=640&ssl=1)
Soluções da equação:
ou
.
![a=0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_ded681eaa02d11064c9a469dd1b3e04c.gif?w=640&ssl=1)
![(3a)^{2015}=1 \Rightarrow 3a=1 \Rightarrow a=\dfrac{1}{3}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_66997797c2318f5745c7ee8ab952782a.gif?w=640&ssl=1)
Soluções do sistema:
e
.
![(0,0,0,0)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_c8e2568b1b465cdc4ee105c6df50dd72.gif?w=640&ssl=1)
![(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_cc15256b48d02696e173666201e4f2f8.gif?w=640&ssl=1)
AVANÇADO:
Primeiramente iremos provar que
é injetiva. Vejamos que se:
![f](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_8fa14cdd754f91cc6554c9e71929cce7.gif?w=640&ssl=1)
![f(m)=f(n) \Rightarrow f(m)+f(n)=f(n)+f(n) \Rightarrow f(f(m)+f(n))=f(f(n)+f(n)) \Rightarrow m+n = n+n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_7d7a54d98d19046f1b90ea17f76f8811.gif?w=640&ssl=1)
o que implica
. Se
então:
![m=n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_a193ebc083b4745370f6f1343383d9cc.gif?w=640&ssl=1)
![k<n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_073c8e74ad8938f323c7b7bf2b67c8c2.gif?w=640&ssl=1)
![f(f(m+k)+f(n-k))=m+n=f(f(m)+f(n))](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_ccfa99740cefe3698b4446f208b350ea.gif?w=640&ssl=1)
Como f é injetiva temos:
![f(m+k)+f(n-k)=f(m)+f(n)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_2d49b35848d0e4c15d3f8b1cc4404b74.gif?w=640&ssl=1)
![m,n \in \mathbb{N}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_44384f4efee305d88d2a708ad3eb791c.gif?w=640&ssl=1)
![k<n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_073c8e74ad8938f323c7b7bf2b67c8c2.gif?w=640&ssl=1)
Suponha se possível que
1" />. Então
, mas vejamos que:
![f(1)=b data-recalc-dims=](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_a95d9eef2e5bf6ee67750fc75f150ffa.gif?w=640&ssl=1)
![b \ge 2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_6481e931773f85782007d7ad3e266fd6.gif?w=640&ssl=1)
![f(2b)=f(f(1)+f(1))=2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_52c1cb7b31521214e55cf1b563749666.gif?w=640&ssl=1)
e,
![f(b+2)=f(f(1)+f(2b))=1+2b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_f2b6493dc2e9caeb0f74dd647cca6c8b.gif?w=640&ssl=1)
Se
então
e
ao mesmo tempo pela relação acima, portanto temos um absurdo. Logo
é maior que
. Agora por (*) temos:
![b=2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_19bf9442bea375a24abb4c22e9951a92.gif?w=640&ssl=1)
![f(4)=2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_e65b2246d5eff524b97be1ba18fd023d.gif?w=640&ssl=1)
![f(4)=5](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_00ec93081a69018fffccd7b74a9e3e9e.gif?w=640&ssl=1)
![b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.gif?w=640&ssl=1)
![2](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_c81e728d9d4c2f636f067f89cc14862c.gif?w=640&ssl=1)
![f(2b)+f(1)=f(2b-(b-2))+f(1+b-2)=f(b+2)+f(b-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_9db14af4074aec38eddb25993332b09f.gif?w=640&ssl=1)
Isso implica
, o que nos fornece
, mas
e
0" />, portanto
. Logo:
![2+b=1+2b+f(b-1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_f72fd84da6ec08bdc2e054ad9830d422.gif?w=640&ssl=1)
![f(b-1)=1-b](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_ec63bbfd4129d78ecbeae2c44147e08e.gif?w=640&ssl=1)
![1-b<0](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_72100de7c6135d44bdac51e268cf8c80.gif?w=640&ssl=1)
![f(b-1) data-recalc-dims=](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_208d00c39980d1f3f8870c8695d03477.gif?w=640&ssl=1)
![b=1](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_3c94d884933477acdc14fc70da4b987a.gif?w=640&ssl=1)
![2=f(2b)=f(2)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_609fd011835b563ab11a5a82410342dd.gif?w=640&ssl=1)
Por indução, suponha que
para todo
, então agora use
![f(k)=k](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_8244a463fca7342ef08f1933b6e20e55.gif?w=640&ssl=1)
![k \le n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_e9615f6cdf98f0dfd3ee2f2113fec52e.gif?w=640&ssl=1)
![n+1=f(f(n)+f(1))=f(n+1)](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_644e546c98bdc5ef8650bedf73611a09.gif?w=640&ssl=1)
logo concluímos que
para todo
.
![f(n)=n](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_3c21a3db5930a60e91aaa3b04fbf094a.gif?w=640&ssl=1)
![n \in \mathbb{N}](https://i0.wp.com/noic.com.br/wp-content/plugins/latex/cache/tex_d20de1fa124517c91ebb375b63ec56ee.gif?w=640&ssl=1)