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Question :

Solve for natural number the equation :

$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

My try :

Let : $X=5^{x}$ and $Y=2^{y}$ so above equation

equivalent :

$2X^{2}+(Y-4)X-6Y^{2}-Y+2=0$

We solve this equation for $X$

$\Delta =(7Y)^{2}$ mean : $X_{1}=\frac{3}{2}Y+1$ and $X_{2}=1-2Y$

From here how I can find $X$ and $Y$ , this is

all my effort ?

Thanks!

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    $\begingroup$ Please use \cdot to get a centered dot for multiplication instead of periods. Also you probably meant $Y=2^y$ $\endgroup$ Oct 6, 2019 at 21:41
  • $\begingroup$ I believe your term $6Y^2$ should be $6Y$. $\endgroup$ Oct 6, 2019 at 21:45
  • $\begingroup$ See again sir @Ross Millikan , I multiple equation $×2$ in first $\endgroup$ Oct 6, 2019 at 21:52
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    $\begingroup$ Yes, I saw that, but multiplying $3\cdot 2^y$ by $2$ gives $6Y$ $\endgroup$ Oct 6, 2019 at 21:53
  • $\begingroup$ Ooh , yes see I add something thank you very much sir 😍 $\endgroup$ Oct 6, 2019 at 21:56

3 Answers 3

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Different substitutions, let $$ u = 5^x \; , \; \; v = 2^{y-1} $$
Then $$ \frac{1}{4} \left( (2u+v-2)^2 - 49 v^2 \right) = 0 \; , \; \; $$ $$ \left( 2u+8v-2 \right) \left( 2u-6v-2 \right) = 0 \; , \; \; $$ $$ \left( u+4v-1 \right) \left( u-3v-1 \right) = 0 \; , \; \; $$ Both $u,v > 0$ so we are left with $$ 5^x = 3 \cdot 2^{y-1} + 1 $$ For example, $3 \cdot 8 + 1 = 25.$

Monday: in some cases there is an elementary proof that we have already found the largest solution. We already have $3 \cdot 8 + 1 = 25.$ If we were to have a larger solution, it would be of the form $$ 24 \cdot 2^s + 1 = 25 \cdot 5^t $$ Subtract 25 from both sides, $$ 24 \cdot 2^s - 24 = 25 \cdot 5^t - 25 \; , \; $$ $$ 24 \left( 2^s - 1 \right) = 25 \left( 5^t - 1 \right) $$ We will ASSUME $s,t \geq 1$ and get a contradiction.

Since $25 | (2^s - 1),$ we find $2^s \equiv 1 \pmod {25}$ and then $s$ is divisible by 20. This $$ 2^{20} - 1 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41. $$ The actual $2^s - 1$ is divisible by $2^{20} - 1,$ and therefore by the prime $41.$

Since $41 | (5^t - 1),$ we find $5^t \equiv 1 \pmod {41}$ and then $t$ is divisible by 20. This $$ 5^{20} - 1 = 2^4 \cdot 3 \cdot 11 \cdot 13 \cdot 41 \cdot 71\cdot 521 \cdot 9161. $$ The actual $5^t - 1$ is divisible by $5^{20} - 1,$ and therefore by $2^4 = 16.$

We have arrived at $$ 16 | 24 (2^s - 1). $$ Divide throsgh by $8,$ we get $$ 2 | 3 (2^s - 1) $$ This is a contradiction, as both $3$ and $2^s - 1$ are odd when $s \geq 1.$ The contradiction tells us the assumptions are wrong, and $$ s = 0 $$ This completes the proof that $3 \cdot 8 + 1 = 25$ is the largest solution.

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Now substitute back: you get that $5^x=3\cdot 2^{y-1}+1$ or $5^x=1-2^{y+1}$. The second one is impossible cause the RHS is $\leq 0$. For the first one to hold, looking mod 3 you see that $x$ needs to be even, so write $x=2x'$. Then $(5^{x'}-1)(5^{x'}+1)=3\cdot 2^{y-1}$. So the only chance is that $5^{x'}-1=2^{y_1}$ and $5^{x'}+1=3\cdot 2^{y_2}$ for some $y_1,y_2\in \mathbb N$, or the other way round. Substract term by term and you'll see immediately that the minimum of $y_1,y_2$ has to be either $0$ or $1$. Now you can check by hand case by case.

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$5^{2x}-3\cdot 2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

$(5^{2x}-2\cdot 5^x + 1)+ (5^{x}2^{y-1}-2^{y-1})-3\cdot 2^{2y}=0$

$(5^x-1)^2 + 2^{y-1}(5^x - 1)-3\cdot 2^{2y}=0$

If we solve for $5^x-1$ we get

$5^x-1 =\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 12*2^{2y}}}{2}=$

$\frac {-2^{y-1} \pm\sqrt{2^{2y-2} + 48*2^{2y-2}}}{2}=\frac {-2^{y-1} \pm\sqrt{49*2^{2y-2}}}{2}=$

$\frac {-2^{y-1} \pm 7*2^{y-1}}{2}=3*2^{y-1}$

Or $5^x = 3*2^{y-1} + 1$.

$5^x = 2^y + 2^{y-1} + 1$

But $5^x = (4+1)^x = \sum_{j=0}^x C_{x,j} 4^j$

We have a solution for $x=2$ and $y=4$ ($5^2 = 3*2^3+1$) which seems to work as ${2\choose 1}4 + 4^2 = 2*2^2 + 2^4= 2^3+2^4$. But in general it doesn't seem like a likely occurrence.

For this to occur we need $\sum_{j=1}^x C_{x,j} 4^j = 2^{y-1} + 2^y$.

There's probably a really easy way to prove this is impossible if $x > 2$. If we divide the LHS by $4$ we get a remainder of $x\pmod 4$ and the RHS is $0\pmod 4$ if $y> 4$.... I think we have some chasing involved. $2^{y-2}|x$ but then $5^x$ is huge compared to $3*2^{y-1}+1$....

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  • $\begingroup$ I added a proof. There is an elementary method, given primes $p,q$ and integers $a,b$ and positive integer variables $x,y,$ of attempting to show that the largest solution to $a(p^x - 1) = b (q^y - 1)$ has already been found. $\endgroup$
    – Will Jagy
    Oct 7, 2019 at 18:17

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