Looking at $\pmod{7}$ we see that $a^3+b^3+1 \equiv 0\pmod{7}$ so one of $a$ or $b$ is $7$ and we just check two cases.

Write $20a^3 = (b+1)(b^2-b+1)$. Now note that $b^2-b+1 \equiv 3 \mod b+1$. So $d=(b^2-b+1,b+1) = (3,b+1) = 3$ or $1$. If $d=3$, then $3 \mid 20a^3 \longrightarrow a=3$, which doesn't satisfy. So $d=1$. Then $a^3$ divides exactly one of $\{ b+1,b^2-b+1 \}$. If $a^3 \mid b+1$, then $$\frac{20}{b^2-b+1} = \frac{b+1}{a^3} \geq 1 \longrightarrow 20 \geq b^2-b+1 \geq b+1 \longrightarrow b \leq 19$$If $a^3 \mid b^2-b+1$, then $$\frac{20}{b+1} = \frac{b^2-b+1}{a^3} \geq 1 \longrightarrow 20 \geq b+1 \longrightarrow b \leq 19$$So we will just check $b=2,3,5,7,11,13,17,19$. Since $b^3 + 1 \equiv 0 \mod 20$, we can easily eliminate all of them except $b=19$ which gives a solution $(7,19)$

Write following equation like $20a^3=b^3+1=(b+1)(b^2-b+1).$ Now we will have cases where we can find prime numbers that satisfy the equation. 1st case where b can be a prime number is: $b+1=4$ $b=3$, but $5a^3=7$ and $a$ is not prime number. 2nd case where b can be a prime number is: $b+1=20$ $b=19$ and $a^3=343$, so $a=7$, both are prime and that is one solution 3rd case is: $b+1=4a^2$ and $b^2-b+1=5a$, where $a$ and $b$ are not even natural numbers, so the only solution is $(a,b)=(7,19)$

My Solution: $20p^{3}-q^{3}=1\Rightarrow 20p^{3}=(q+1)(q^{2}-q+1)$ Note,that $gcd(q+1,q^{2}-q+1)=1$ or $3$.So we have two cases: Case 1:$gcd(q+1,q^{2}-q+1)=1$ So $20\mid (q+1)(q^{2}-q+1)\Rightarrow 20\mid q+1$ or $20\mid q^{2}-q+1$,but $q^{2}-q+1\equiv 1(mod2)$,so $20$ can't divide an odd number. So we have $20\mid q+1\Rightarrow 20k=q+1;k\in Z^{+}$ $\Rightarrow 20p^{3}=20k(q^{2}-q+1)\Rightarrow k\mid p^{3}\Rightarrow k=1,p,p^{2},p^{3}$ If $k=1\Rightarrow 20=q+1\Rightarrow q=19,p=7$ For the other cases,we have that $p\mid 60\Rightarrow p=2,3,5$,which gives no solution for $q$. Case 2:$gcd(q+1,q^{2}-q+1)=3$ $\Rightarrow 9\mid 20p^{3}\Rightarrow 9\mid p^{3}\Rightarrow p=3\Rightarrow 539=q^{3}\Rightarrow q\notin N$ And finally,the only solution is:$(p,q)=(7,19)$ $Q.E.D$

@above: you're letting out many cases... Rewrite as $20a^3=(b+1)(b^2-b+1) $. Notice $\gcd (b+1,b^2-b+1)=\gcd (b+1,3) $. If both factors are relatively primes. $\textbf{Case 1.}$ $a\mid b+1$ So $a^3\le b+1\le b^2-b+1\le 20$ so there are no solutions. $\textbf {Case 2.} $ $a\mid b^2-b+1$ So $a^3\le b^2-b+1\le 373=7^3$ this give us the solution $(a,b)=(7,19)$. If both factors are divisible by $3$ we have $9\mid 20a^3\implies a=3$ which isn't a solution.

I have very simple solution to this problem. Rewrite equation in $$21a^3-(a^3 + b^3 + 1) = 0$$Which means that $$a^3 + b^3 + 1\equiv 0 \pmod 7$$And we know that cube of integer can have remainders $(0, 1, -1)$ when divided by $7$. Thus one $a$ or $b$ is divisible by $7$ so one is equal $7$. Case 1: $a = 7$ We get that $b^3 = 20*7^3 - 1$ thus $b = 19$. Case 2: $b = 7$ We get that $a$ is not integer so there is no solution. Thus only solution is $(a, b) = (7, 19)$.

Note, that $21a^3-a^3-b^3-1=0$, hence, $a^3+b^3+1 \equiv 0\pmod{7}$, note that then, $a^3+b^3 \equiv -1\pmod{7}$, hence, one of $a,b$ is divisible by $7,$ we will take cases. Case 1: $a=7.$ $b^3=20\cdot 343-1=6859=19^3$, hence, $(a,b)=(7,19).$ Case 2: $b=7.$ Note, that $a^3=\frac{344}{20}$, which is not an integer, thus no solutions in this case. So, the answer, is $\boxed{(7,19)}.$

$20p^3=(q+1)(q^2-q+1)$ $\gcd(q+1,q^2-q+1)=1$ $q+1 $is even $q^2-q+1$ is odd $p$ is odd 1.$p=5$ contradicition. 2.So $q+1=p^3,20,4$ $q=19,p=7$

taking everything mod $7$ gives us $a^{3}+b^{3}=6 (\text{mod } 7)$. Testing out cubes mod $7$, we can conclude one of $a$ or $b$ must be divisible by $7$, or $7$ itself. Testing out some cases, the only solution is $a=7, b=19$

Here is my solution We know that $x\equiv 0,1,2,3,4,5,6\pmod{7} \implies x^3\equiv 0,1,6\pmod{7} \implies 20x^3\equiv 0,1,6\pmod{7} \implies$ $20a^3\equiv 0,1,6\pmod{7}$ $...(1)$ $b^3\equiv 0,1,6\pmod{7}$ $...(2)$ We subtract $(1)$ and $(2)$ and we get: $20a^3-b^3\equiv 0,1,2,5,6\pmod{7}$ Combining with the condition we get: $1\equiv 0,1,2,5,6\pmod{7}$ but we know that $1\equiv 1\pmod{7}$. So we have two cases Case 1. $20a^3\equiv 1\pmod{7}$ , $b^3\equiv 0 \pmod{7}$ From $b^3\equiv 0 \pmod{7} \implies b\equiv 0 \pmod{7} \implies 7\mid b \implies b=7$ We plug this in and we have: $20a^3-b^3=1 \implies 20a^3-7^3=1 \implies 20a^3-343=1 \implies 20a^3=344$ we can clearly see that for $a$-we have no solution Case 2. $20a^3\equiv 0\pmod{7}$ , $b^3\equiv 6 \pmod{7}$ From $20a^3\equiv 0\pmod{7} \implies a^3\equiv 0\pmod{7} \implies a\equiv 0\pmod{7} \implies 7\mid a \implies a=7$ We plug this in and we have: $20a^3-b^3=1 \implies 20*7^3-b^3=1 \implies 6859=b^3 \implies 19=b \implies b=19$ So $(a,b)=(7,19)$