Solve $a^2+b^n=c^2$

Question

Let $a,b,c$ be co-prime integers >1, for all $n>2$, I need help finding the integral solutions of the diophantine equation $a^2+b^n=c^2$. I saw the result but I am curious about to how to get there. Thanks.

Answer 1

I think it could be solved as follows. Note that we have $$ c^2 - b\bigl(b^{(n-1)/2}\bigr)^2 = a^2. $$ Now $$(c,b^{(n-1)/2},a)=\biggl(\frac{r^2+bs^2}{t},\frac{2rs}{t},\frac{r^2-bs^2}{t}\biggr)$$ for some integers $r,s,t$ with $t \ne 0$.

Answer 2

Very simply: $$a^2+b^{n}=c^2$$

Degradable trên multipliers. $$c^2-a^2=(c-a)(c+a)=b^{n}=qt$$

Solutions obtained: $$a=\frac{t-q}{2}$$ $$c=\frac{t+q}{2}$$

It's easier. Can a formula in another form to write, but the expression is long and still nobody wants. You can write such a formula.

$$a=k^{n}(2(p-s))^{n-2}-(p-s)^2$$

$$b=2k(p-s)$$

$$c=k^{n}(2(p-s))^{n-2}+(p-s)^2$$