(3^{p-1} - 1)/p is a perfect square for prime p

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(3^{p-1} - 1)/p is a perfect square for prime p

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Source: 2017 Saudi Arabia JBMO TST 1.2

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#1 h May 28, 2020, 4:10 AM

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Find all prime numbers $p$ such that $\frac{3^{p-1} - 1}{p}$ is a perfect square.

This post has been edited 1 time. Last edited by parmenides51, May 28, 2020, 4:13 AM

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#2 h May 28, 2020, 4:15 AM

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Test for small prime numbers and see that like after $p=2,5$ there are no more perfect squares - at least I didn't find anymore perfect squares. Then try to prove it?

This post has been edited 1 time. Last edited by Unidentifiable, May 28, 2020, 4:15 AM

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#3 h May 28, 2020, 7:54 AM

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parmenides51 wrote:

Find all prime numbers $p$ such that $\frac{3^{p-1} - 1}{p}$ is a perfect square.

Easy one

Clearly $p=2$ is a solution, suppose that $p >2$ , then $p-1=2a$ giving that
$(3^a-1)(3^a+1) = 3^{p-1} = pC^2$ Since $\gcd(3^a-1,3^a+1)=2$ , then
$\textbf{Case.1}$
$3^a-1 = 2px^2, 3^a+1 = 2y^2$ which is absurd because $0 \equiv 3^a \equiv 2y^2 - 1 \equiv 1 \not\equiv 0 \pmod3$ . $\blacksquare$
$\textbf{Case.2}$
$3^a-1 = 2x^2, 3^a +1 = 2py^2$ Clearly $a$ is even. Indeed, if $a$ was odd then $4 \mid 3^a + 1$ giving that $2 \mid y$ so $8 \mid 3^a +1$ , a contradiction.
Let $a=2b$ , then $(3^b - 1)(3^b+1)=2x^2$ ..
As previously, $\gcd(3^b-1,3^b+1)=2$ but $3^b +1 \neq 2r^2$ , Thus $3^b +1 = r^2$ giving that
$\Rightarrow 3^b = (r-1)(r+1)$ So $r-1=3^h$ and $r+1 = 3^t$ giving $3^t - 3^h = 2$ , and it is easy to see that $(t,h) =(1,0)$ , hence $p=2a+1=4b+1 = 4(t+h)+1 =5$ , we are done. $\blacksquare$

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#4 h Apr 14, 2025, 10:16 AM

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a question similar to this came in INMO

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