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The solutions of $a^2 = b^d -3^c$ are in the form $(a, b, c, d) = ((46)27^t, (13)9^t, 6t+4, 3)$. This is done by using calculator. As per my calculator, I have checked some terms, which are satisfied the above cited equation. Now, my question is, is there any other forms of solutions? if there, how to examine?

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$(1,2,1,2)$, $(1,4,1,1)$. –  Gerry Myerson Oct 20 '12 at 6:40
    
@GerryMyerson!you are really helping me always. But, I would like to learn the way to find solutions. Otherwise, my life long duty to upload only questions. please try to understand me and tell me, how to solve such equations mathematically. –  VMRFDU Oct 20 '12 at 6:50
    
The way I found those two solutions was, I saw that no matter what $a$ and $c$ are, I can always let $d=1$ and solve for $b$. So I took $a=c=1$, found $b^d=4$, which gave two solutions. –  Gerry Myerson Oct 20 '12 at 9:01
    
@GerryMyerson!this is somewhat good. –  VMRFDU Oct 20 '12 at 9:25
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1 Answer

One infinite, albeit somewhat trivial, family of solutions has $d=2$. This leads to $b^2-a^2=3^c$. It is well-known, how to express a number as a difference of two squares. For any $r\lt c/2$, we get $b=(3^r+3^{c-r})/2$, $a=(3^{c-r}-3^r)/2$.

Another solution is given by $10^2=7^3-3^5$.

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Gerry Myerson! In my post I showed a, b, c and d values in terms of t by fixing d = 3. You have told that, in terms of r by fixing d = 2. Now, tell me if d = odd prime greater than or equal to 5, what type of solutions we can expect? –  VMRFDU Oct 20 '12 at 9:24
    
! without taking d= some fixed number, can we get a, b, c and d values in terms of single variable, which I have shown in the post. I need in single variable with proof. I am not looking cases. –  VMRFDU Oct 20 '12 at 9:32
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