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My motivation is my try for solving the following great question and then I stuck:

Find all positive integer solutions of $(a,b,n)$ s.t. $(2^a-1)(2^b-1)=n^2$

so any variable in this question $\in \{1,2,3,...\}$

$(2^a-1)(2^b-1)=n^2$

first let $a=1$

then:

$(2-1)(2^b-1)=(2^b-1)=n^2$

If $n$ is positive integer then:

$ n \equiv 0,\pm1,\pm2 \pmod {5} $

then: $ n^{2} \equiv 0,1,4 \pmod {5} $

and

$ (2^{b}-1) \equiv 0 \pmod {5} $ when b=4m

$ (2^{b}-1) \equiv 1 \pmod {5} $ when b=4m-3

$ (2^{b}-1) \equiv 2 \pmod {5} $ when b=4m-1

$ (2^{b}-1) \equiv 3 \pmod {5} $ when b=4m-2

So probable solutions are when $b=4m,4m-3$

so I made the following equations:

First Case: let $b=4m, n=(10k-5)$ because $n^2 \equiv 0 \pmod 5$ for all k

$2^{4m}-1=(10k-5)^2$

Second Case: let $b=4m-3,n=(10k-9)$ because $n^2 \equiv 1 \pmod 5$ for all k

$2^{4m-3}-1=(10k-9)^2$

Third Case : let $b=4m-3,n=(10k-1)$ because $n^2 \equiv 1 \pmod 5$ for all k

$2^{4m-3}-1=(10k-1)^2$

*note that $n=(10k-3)^2$ is included in Third Case: for example $n^2=((20-3)^2)^2=17^4=(280+9)^2$

Are there solutions or not ?

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    $\begingroup$ $\left(2^{2}-1\right)\left(2^{2}-1\right)=3^{2}$ $\endgroup$
    – Mason
    Sep 12 at 6:29
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    $\begingroup$ @Mason Thank you but I am trying to find solutions other than when $a=b$ $\endgroup$ Sep 12 at 6:31
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    $\begingroup$ The question in the title is trivial. The right side is a square, the left side is one less than a square, so you're asking for two squares to differ by one, and the only way that happens is for the squares to be zero and one. $\endgroup$ Sep 12 at 7:50
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    $\begingroup$ The other two cases, just look at them modulo eight. $\endgroup$ Sep 12 at 12:14
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    $\begingroup$ Looks good to me. $\endgroup$ Sep 12 at 13:02
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Depending in comment by @Gerry Myerson ,many thanks for him.

For the first equation:

$2^{4m}-1=(10k-5)^2$ it turns out to be trivial because squares of integers can't differ by one.

For the second equation:

$2^{4m-3}-1=(10k-9)^2$

We will do it by modulo $8$

If n is positive integer then:

$n \equiv 0,\pm1,\pm2,\pm3,4 \pmod 8$

then: $n^{2} \equiv 0,1,4 \pmod 8$

$(2^{4m-3}-1) \equiv 1 \pmod8$ for $m=1$

$(2^{4m-3}-1) \equiv 7 \pmod8$ for $m \in\{2,3,4,...\}$

so $m=1$ is probable

so the equation $2^{4m-3}-1=(10k-9)^2$ becomes $1=(10k-9)^2$ so k=1

then : $a=1=b,n=1$ which lies under the solution set $a=b$

Now for the third equation

$2^{4m-3}-1=(10k-1)^2$

we know from above that

$n^{2} \equiv 0,1,4 \pmod 8$

and

$(2^{4m-3}-1) \equiv 1 \pmod8$ for $m=1$

$(2^{4m-3}-1) \equiv 7 \pmod8$ for $m \in\{2,3,4,...\}$

so $m=1$ is probable

so the equation $2^{4m-3}-1=(10k-1)^2$ becomes $1=(10k-1)^2$ so there is no $k$ satisfies this equation since I assumed $k \in \{1,2,3,...\}$.

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