Find all solutions to 3^a - 1 = 2b^2 - B.M.O. problem

[al-mahed]

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 $a = 2k+1$, we have $3^{2k+1} - 1 = 2b^2 ==> 3^{2k+1} - 3 = 2b^2-2 ==>3(3^{2k} - 1) = 2(b^2-1)$
so
$3(3^k - 1)(3^k+1) = 2(b-1)(b+1)$

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 $gcd(3^k - 1,\:3^k+1) = 2 ==> (3^k - 1)(3^k+1)\equiv 0\\: mod\\: 8 ===>$ b is odd $==> b = 2n+1 ==>$

$b^2 = 4n(n+1)+1 ==>\: 3^{2k+1} - 1 = 2[4n(n+1)+1] ==>\: 3^{2k+1} = 8n(n+1)+3 ==>$

$==>\: 3^{2k} = (3^k)^2=\frac{8n(n+1)}{3}+1$ since $(3^k)^2$ is a perfect odd square, then

$(3^k)^2 =\frac{8m(m+1)}{2}+1 =\frac{8n(n+1)}{3}+1==>\: 3m(m+1)=2n(n+1)$ 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 $$\frac{n(n+1)}{3}$$ cannot be a triangular number, but it is so obvious that I cannot solve it in a simple manner.

 

[robert Ihnot]

The case of a=5, gives b=11. We then have, k=2, n=5, which fulfills 3^(k^2) = 8n(n+1)/3 +1, or 81= 8*5*6/3+1 or 81= 80+1.

For triangle number 10, we have 10 = 4*5/2 = 5*6/3.

 

[al-mahed]

The case of a=5, gives b=11. We then have, k=2, n=5, which fulfills 3^(k^2) = 8n(n+1)/3 +1, or 81= 8*5*6/3+1 or 81= 80+1.

For triangle number 10, we have 10 = 4*5/2 = 5*6/3.

Yes, you're absolutely correct! This small example skipped me. But is it the only one? Perhaps there is another easier way to find the solution.

In fact $$\frac{n(n+1)}{3}$$ can be triangular infinitely many times, what I was trying to say is that when it is triangular the square $$\frac{8n(n+1)}{3}+1$$ is not only divisible by 3, or maybe it is not divisible by 3 at all (perhaps 10 is the only solution).

I think this way to prove it is too complicated.

thank you for your imput

 

[snipez90]

Unfortunately trying small cases doesn't really uncover a pattern (which makes this a olympiad problem I guess). For diophantine equations, if using clever factorizations or inequalities to force solutions doesn't work, then perhaps try some more advanced tools. You could try converting this to a "[URL equation[/URL] and consider a convenient mod to uncover solutions. I haven't tried that yet, but I haven't found an elegant way to do this.

 

[al-mahed]

Thank you for your hints, in fact I was trying a different strategy now:

lets write the equation as $$\frac{3^a-1}{3-1}=b^2$$

we already demosntrated that if a=2k+1 then b=2n+1, so

==> $$\frac{3^{2k+1}-1}{3-1}=1+3+9+...+3^{2k}=(2n+1)^2=4n(n+1)+1$$ ==>

==> $$1+3+9+...+3^{2k}=4n(n+1)+1 \ ==> \ 3+9+...+3^{2k}=4n(n+1)$$

there are 2k terms on the sum at the left

rewriting

$$(3+9)_1+(27+81)_2+...+(3^{2k-1}+3^{2k})_k=4n(n+1)\ ==>$$

$$\ ==> \frac{(3+9)_1}{4}+\frac{(27+81)_2}{4}+...+\frac{(3^{2k-1}+3^{2k})_k}{4}=n(n+1)$$

if we notice that $$3^r+3^{r+1}=3^r(3+1)=4*3^r$$ then we can write the above as

$$\ ==> \frac{3^1(1+3)_1}{4}+\frac{3^3(1+3)_2}{4}+...+\frac{3^{2k-1}(1+3)__k}{4}=n(n+1)$$

cancel 1+3 with 4, we have

$$(3^1)_{1}+(3^3)_{2}+...+(3^{2k-1})_{k}=n(n+1)$$

we notice that n(n+1) is even, so as there are k terms on the sum at the left, and the terms are all odd, then k is even, because an odd quantity of odd numbers added is an odd number, so we must have an even number of terms, so k is even, let's call k = 2x

from here, we conclude that as a = 2k+1 = 4x+1, obviously a is of the form 4x+1 and we can factor the equation as follows

$$3^{4x+1}-1=2b^2 \ ==> \ 3^{4x+1}-3=2b^2-2 \ ==> \ 3(3^{4x}-1)=2(b^2-1) \ ==>$$

$$\ ==> \ 3(3^{2x}-1)(3^{2x}+1)=2(b-1)(b+1) \ ==>$$

$$\ ==> \ 3(3^x-1)(3^x+1)(3^{2x}+1)=2(b-1)(b+1)=2(2n)(2n+2)=8n(n+1)$$

as $$(3+9)_1+(27+81)_2+...+(3^{2k-1}+3^{2k})_k=4n(n+1)$$ therefore

$$\ ==>\ 2[(3+9)_1+(27+81)_2+...+(3^{2k-1}+3^{2k})_k]=3(3^x-1)(3^x+1)(3^{2x}+1)$$

arranging the terms at the left properly, and noticing that 2 = 3^1-1

$$(3^1-1)[3^1(1+3^1)_{1}+3^3(1+3^1)_{2}+...+3^{2k-1}(1+3^1)_{k}]=3(3^x-1)(3^x+1)(3^{2x}+1)$$
$$(3^1-1)(3^1+1)[3^1_{1}+3^3_{2}+...+3^{2k-1}_{k}]=3(3^x-1)(3^x+1)(3^{2x}+1)$$

as k is even = 2x, we can line up the powers of 3 two-by-two again, and I'll now insert 2x where there is k written

$$(3^1-1)(3^1+1)[(3^1+3^3)_{1}+(3^5+3^7)_{2}+...+(3^{4x-3}+3^{4x-1})_{x}]=3(3^x-1)(3^x+1)(3^{2x}+1)=$$

$$=(3^1-1)(3^1+1)[3^1(1+3^2)_{1}+3^5(1+3^2)_{2}+...+3^{4x-3}(1+3^2)_{x}]=3(3^x-1)(3^x+1)(3^{2x}+1)=$$

$$(3^1-1)(3^1+1)(3^2+1)[3^1_{1}+3^5_{2}+...+3^{4x-3}_{x}]=3(3^x-1)(3^x+1)(3^{2x}+1)=$$

$$3(3^1-1)(3^1+1)(3^2+1)[3^0_{1}+3^4_{2}+...+3^{4x-4}_{x}]=3(3^x-1)(3^x+1)(3^{2x}+1)$$

so we have x terms there, the equality suggests that x = 1, but I cannot see how to proceed from here, any ideas?

best regards