# Solutions of $a^2 = b^d -3^c$

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

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 $10^2=7^3-3^5$.