Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
$(1,2,1,2)$, $(1,4,1,1)$. – 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 $10^2=7^3-3^5$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.