Find all $(a,b)\in\Bbb{N},$ such that $5^a +2^b +8$ is a perfect square.
My approach: Let $5^a +2^b +8=k^2\implies \bigg(k+5^{\frac{a}{2}}\bigg)\bigg(k-5^{\frac{a}{2}}\bigg)=8\bigg(1+2^{b-3}\bigg).$ The RHS is greater than zero and even,so each of $\bigg(k+5^{\frac{a}{2}}\bigg)$ and $\bigg(k-5^{\frac{a}{2}}\bigg)$ is even and $k$ is odd. $1+2^{b-3}$ is even only when $b=3.$ Case(1): When $b=3,k^2 =5^a +16;$ Let $k=2m+1\implies 4m^2 +4m+1=5^a +16$ or $4m(m+1)=5^a +15=5\bigg(5^{a-1}+3\bigg).$ If $m$ is odd,$m+1$ is even and vice-versa. Equating odd and even parts , Sub-case(a): $m+1=5\implies m=4,$ and $4m=16=5^{a-1}+3\implies 5^{a-1}=13\implies$ no natural number exist. Sub-case (b): $m=5$ and $4(m+1)=24=5^{a-1}+3\implies 5^{a-1}=21\implies$ again no natural number exist. So,there is no natural number solution set exist for $b=3.$ This implies that $1+2^{b-3}$ is odd. This means that even components: $k+5^{\frac{a}{2}}=c\bigg(1+2^{b-3}\bigg).$ And $8=8\bigg(k-5^{\frac{a}{2}}\bigg)......(A).$ where $c$ is even . From equation $(A),$ $c$ has to be non-zero even factor of $8,$ but not $8$ because otherwise $k-5^{\frac{a}{2}}=1$ and odd. So, $c=(2,4).$ Case(2): When $c=2$ $k-5^{\frac{a}{2}}=4$ $k+5^{\frac{a}{2}}=2\bigg(1+2^{b-3})\implies 5^{\frac{a}{2}}=2^{b-3}-1$ Or $2^{b-3}-5^{\frac{a}{2}}=1.$ The only $a$ and $b$ for which this works is $a=2$ and $b=4\implies k=7.$ There are other solutions too!
And I have no clue how to proceed from here,also if any other way is possible,then please try to show here. Thank you for each and everyone trying to indulge in solving it!
Is there any complete solution?
Equation: $2^b + 5^a + 8 = N^2$ where $a$, $b$ and $N$ are some natural numbers.
1. Case $a = 0$: For $a = 0$, the equation becomes $2^b + 9 = N^2$. $2^b = (N+3)(N-3)$. Let $2^{k2} = N+3$ and $2^{k1} = N-3$, where $k_2 > k_1$ and $k_1 + k_2 = b$. $2^{k2} - 2^{k1} = 6$. This implies $2^{k1}(2^{k2-k1}-1) = 2 \times 3$. Forced by this, $k_1 = 1$ and thus $N = 5$. Consequently, $2^b = 16$ and $b = 4$, with $k_2 = 3$. Therefore, $(a, b) = (0, 4)$.
2. Case $a = 1$: For $a = 1$, the equation becomes $2^b + 13 = N^2$. Note that $N$ must be odd, and $N^2 \equiv 1 \pmod 8$. Therefore, $2^b + 13 \equiv 1 \pmod 8$, leading to $2^b \equiv 4 \pmod 8$. Thus, $b = 2$ and $17 = N^2$ which leads to a contradiction. Hence, there are no integer solutions for $a = 1$.
3. Case $a = 2$: For $a = 2$, the equation becomes $2^b + 33 = N^2$. It is clear that $N$ must be odd, and hence $2^b \equiv 0 \pmod 8$, implying $b \geq 3$. Considering $2^b \equiv N^2 \pmod 3$, with $N^2 \equiv 1 \text{ or } 0 \pmod 3$ depending on whether $N$ is divisible by 3, and $2^b \equiv 2 \text{ or } 1 \pmod 3$ depending on whether $b$ is odd or even, we conclude that $b$ must be even. Let $b = 2p$, and the equation becomes $33 = (N+2^p)(N-2^p)$. Applying unique prime factorization, we consider the following cases:
4. Case $a > 2$: For $a > 2$, the equation becomes $2^b + 5^a + 8 = N^2$. Since $N$ must be odd, we have $2^b + 5^a \equiv N^2\equiv 1 \pmod 8$. Notice that $5^a \equiv 5 \text{ or } 1 \pmod 8$ depending on whether $a$ is odd or even, and $2^b \equiv 2, 4, \text{ or } 0 \pmod 8$ depending on whether $b = 1, 2, \text{ or } >2$.
Case 4.1: $a$ is odd and $b = 2.$ The equation becomes $5^a + 12 = N^2$. As $N^2 \equiv 1 \text{ or } 0 \pmod 3$, depending on whether $N$ is divisible by $3$ or not. This contradict with $5^a \equiv 2\pmod 3$ if $a$ is odd.
Case 4.2: $a$ is even and $b > 2.$ By considering modulo $3$, $2^b \equiv 1 \equiv N^2 \pmod 3$ which implies that $b$ is odd and $N \equiv 1 \pmod 6$.
(Here is the incomplete result that can currently be deduced. There were several errors in the previous deductions. Thank you for pointing them out.)
In summary: