I found a pair that satisfies: $ (a,b)=(3,8)$. Someone has a solution.

I have worked with this pro three days and don't have anything new Until someone post a solution , I will try Gaussian numbers

Let $ a^2+b-1=p^k$, $ b^2-a^3-1=mp^k$ => $ b \equiv 1-a^2 (mod p^k)$ => $ b^2 \equiv (a^2-1)^2 \equiv a^4-2a^2+1 (mod p^k)$ => $ a^4-2a^2 \equiv a^3 (mod p^k)$ => $ a^2(a-2)(a+1) \equiv 0 (mod p^k)$ (1) $ (a+b-1)^2 \equiv (a+1-a^2-1)^2 \equiv a^2(a-1)^2 (mod p^k)$ (2) Consider the case $ p>3$ (1),(2),and $ a^2<p^k+1$ imply $ a=2$=>$ b=p^k-3$,where $ p>3,k>0$ We need to consider $ p=2,p=3$,but this case is not easy for me

Fang-jh wrote: Find all pairs of positive integers $ (a$,$ b)$ such that $ a^2 + b - 1$ is a power of a prime number, $ a^2 + b + 1$ divides $ b^2 - a^3 - 1$, but it doens't divide $ (a + b - 1)^2$. Before I start solving the problem, is there anyone who knows if there is a typo in the statement of the problem? Actually I'm not sure about the following: "$ a^2 + b + 1$ divides $ b^2 - a^3 - 1$, but it doens't divide $ (a + b - 1)^2$". Shouldn't it be "$ a^2+b-1$ divides..."?

Actually this is the right one: Find all the pairs of positive integers $ (a,b)$ such that $ a^2 + b +1$ is a power of prime number $ ; a^2 + b + 1$ can divide $ b^2 - a^3 - 1,$ but it can't divide $ (a + b - 1)^2.$ BTW,it's different from this one: http://www.mathlinks.ro/viewtopic.php?p=1806011#1806011 This is a problem in 2003.

litongyang wrote: Find all the pairs of positive integers $ (a,b)$ such that $ a^2 + b + 1$ is a power of prime number $ ; a^2 + b + 1$ can divide $ b^2 - a^3 - 1,$ but it can't divide $ (a + b - 1)^2.$ Let $ a^2 + b + 1 = p^k$ where $ p$ is a prime and $ k$ is a positive integer. We have $ p^k|b^2 - a^3 - 1\equiv(a^2 + 1)^2 - a^3 - 1 = a^4 - a^3 + 2a^2$ and $ p^k\not|(a + b - 1)^2 = (a + p^k - a^2 - 2)^2\equiv(a^2 - a + 2)^2$. Let $ v_p(a) = d$. Since $ a^2 < p^k$, then $ d < \frac {k}2$. Since $ p^k|a^2(a^2 - a + 2)$, then $ v_p(a^2 - a + 2)\ge k - 2d$. But $ p^k\not|(a^2 - a + 2)$ so $ v_p(a^2 - a + 2) < \frac {k}2$. Thus $ \frac {k}2 > k - 2d$, which gives $ 4d > k > 0$, hence $ p|a$. If $ p$ is odd, then $ p$ cannot divide $ a^2 - a + 2$ otherwise $ p$ divides 2. Thus, in case $ p$ is odd, $ \nu_p(a^2 - a + 2) = 0\ge k - 2d$, which is a contradiction to the fact that $ d < \frac {k}2$. Therefore $ p = 2$. Since $ \gcd(a,a^2 - a + 2) = \gcd(a,2)\le2$, then $ \min\{\nu_p(a),\nu_p(a^2 - a + 2)\} = 1$. (i) $ \nu_p(a) = 1$ So $ a\equiv2\pmod4$, which implies $ 4|a^2 - a + 2$, hence $ k > 2$. But $ k < 4d = 4$, so $ k = 3$. We get $ a^2 + b + 1 = 2^3 = 8$, so $ a = 2$ and $ b = 3$. However, in this case, $ a^2 + b + 1 = 8$ divides $ 16 = (2 + 3 - 1)^2$. (ii) $ \nu_p(a^2 - a + 2) = 1$ Since $ p^k|a^2(a^2 - a + 2)$, then $ 2d + 1\ge k$, hence $ d\ge\frac {k - 1}2$. But we also have $ \frac {k}2 > d$, hence we must have $ d = \frac {k - 1}2$ where $ k$ is odd. Let $ k = 2n + 1$. So $ a = 2^nm$ where $ m$ is odd. Since $ a^2 < 2^k$, we get $ 2^{2n}m^2 < 2^{2n + 1}$, so $ m = 1$. So we get $ a = 2^n$, $ b = 2^k - a^2 - 1 = 2^{2n + 1} - 2^{2n} - 1 = 2^{2n} - 1$, it is easy to check that $ (2^n,2^{2n} - 1)$ satisfies the given conditions for all positive integers $ n > 1$ (for $ n = 1$ it's the same as case i). Answer: $ (2^n,2^{2n} - 1)$ for positive integer $ n > 1$.