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al-mahed
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That's a problem I saw at another math forum. It is said to be the last problem of the last phase of the national Brazilian mathematical olympiad. The case "a is even" is solved, but the case "a is odd" is resisting. I'd apreciate if you could give me any hints about it. I wrote the following proof, though I think at the end more steps are needed to give an absolute proof.
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Let [itex]a = 2k+1[/itex], we have [itex]3^{2k+1} - 1 = 2b^2 ==> 3^{2k+1} - 3 = 2b^2-2 ==>3(3^{2k} - 1) = 2(b^2-1)[/itex]
so
[itex]3(3^k - 1)(3^k+1) = 2(b-1)(b+1)[/itex]
of course there's 2 trivial solutions where k = 0 (a=1) and b = 1, or a = 0 and b = 0, and there is no integer k, but let's suppose that k>0
as [itex]gcd(3^k - 1,\:3^k+1) = 2 ==> (3^k - 1)(3^k+1)\equiv 0\\: mod\\: 8 ===>[/itex] b is odd [itex]==> b = 2n+1 ==>[/itex]
[itex]b^2 = 4n(n+1)+1 ==>\: 3^{2k+1} - 1 = 2[4n(n+1)+1] ==>\: 3^{2k+1} = 8n(n+1)+3 ==>[/itex]
[itex]==>\: 3^{2k} = (3^k)^2=\frac{8n(n+1)}{3}+1[/itex] since [itex](3^k)^2[/itex] is a perfect odd square, then
[itex](3^k)^2 =\frac{8m(m+1)}{2}+1 =\frac{8n(n+1)}{3}+1==>\: 3m(m+1)=2n(n+1)[/itex] where m is a given positive integer
that's certainly false, I've tried to prove it but seems to be really tricky
It is obvious that [tex]\frac{n(n+1)}{3}[/tex] cannot be a triangular number, but it is so obvious that I cannot solve it in a simple manner.
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Let [itex]a = 2k+1[/itex], we have [itex]3^{2k+1} - 1 = 2b^2 ==> 3^{2k+1} - 3 = 2b^2-2 ==>3(3^{2k} - 1) = 2(b^2-1)[/itex]
so
[itex]3(3^k - 1)(3^k+1) = 2(b-1)(b+1)[/itex]
of course there's 2 trivial solutions where k = 0 (a=1) and b = 1, or a = 0 and b = 0, and there is no integer k, but let's suppose that k>0
as [itex]gcd(3^k - 1,\:3^k+1) = 2 ==> (3^k - 1)(3^k+1)\equiv 0\\: mod\\: 8 ===>[/itex] b is odd [itex]==> b = 2n+1 ==>[/itex]
[itex]b^2 = 4n(n+1)+1 ==>\: 3^{2k+1} - 1 = 2[4n(n+1)+1] ==>\: 3^{2k+1} = 8n(n+1)+3 ==>[/itex]
[itex]==>\: 3^{2k} = (3^k)^2=\frac{8n(n+1)}{3}+1[/itex] since [itex](3^k)^2[/itex] is a perfect odd square, then
[itex](3^k)^2 =\frac{8m(m+1)}{2}+1 =\frac{8n(n+1)}{3}+1==>\: 3m(m+1)=2n(n+1)[/itex] where m is a given positive integer
that's certainly false, I've tried to prove it but seems to be really tricky
It is obvious that [tex]\frac{n(n+1)}{3}[/tex] cannot be a triangular number, but it is so obvious that I cannot solve it in a simple manner.