Ok, novamente, com 4 reais positivos
1)Sets of 4 positive numbers are made out of each other according
to the following rule: (a, b, c, d) (ab, bc, cd, da).
Prove that in this (infinite) sequence (a, b, c, d) will
never appear again, except when a = b = c = d = 1.
Primeiramente, observe que
Domingos Jr. ([EMAIL PROTECTED]) escreveu:
Daniel S. Braz wrote:
1)Sets of 4 positive numbers are made out of each other according
to the following rule: (a, b, c, d) (ab, bc, cd, da).
Prove that in this (infinite) sequence (a, b, c, d) will
never appear again, except when a = b = c = d = 1.
[EMAIL PROTECTED] wrote:
Domingos Jr. ([EMAIL PROTECTED]) escreveu:
Daniel S. Braz wrote:
1)Sets of 4 positive numbers are made out of each other according
to the following rule: (a, b, c, d) (ab, bc, cd, da).
Prove that in this (infinite) sequence (a, b, c, d) will
never appear again,
Pessoal,
Alguém poderia me dar uma dica na resolução desses aqui?
1)Sets of 4 positive numbers are made out of each other according
to the following rule: (a, b, c, d) (ab, bc, cd, da).
Prove that in this (infinite) sequence (a, b, c, d) will
never appear again, except when a = b = c = d = 1.
tem certeza que o problema 3 não seria:
What is the largest x for which 4^27 + 4^1000 + 4^x equals the square
of a whole number?
Porque esse problema acho que é da olimpiada soviética de 1972, e a
resposta é 1972
On Thu, 10 Mar 2005 11:43:39 -0300, Daniel S. Braz [EMAIL PROTECTED] wrote:
Daniel S. Braz wrote:
Pessoal,
Alguém poderia me dar uma dica na resolução desses aqui?
1)Sets of 4 positive numbers are made out of each other according
to the following rule: (a, b, c, d) (ab, bc, cd, da).
Prove that in this (infinite) sequence (a, b, c, d) will
never appear again, except when
What is the largest x for which 4^27 + 4^1000 + 4^x equals the square
of a whole number?
4^27 + 4^1000 + 4^x = n^2 = (a+b)^2 = a^2 + 2ab + b^2
temos entao dois quadrados perfeitos, onde 4^x = 2ab e onde 4^x = b^2
como queremos o maior x, 4^x = b^2
a^2 + 2ab + b^2 = 4^27 + 4^1000 + 4^x = (2^27 +
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