Base $a$ and $b$ , next level

Clearly, $a > b$ . We must solve $a^2 + a + 7 = b^2 + 8b + 8$ for positive integers $a$ , $b$ .

Reduce the equation to zero: $(a^2 - b^2) + a - 8b - 1 = 0.$ Factor the difference of squares: $(a + b)(a - b) + a - 8b - 1 = 0.$ Write $a - 8b$ also in terms of the sum and difference: $(a + b)(a - b) - \frac72(a + b) + \frac92(a - b) - 1 = 0.$ This invites to a factorization in the form $(a + b + \dots)(a - b + \dots) = \dots$ : $\left(a + b + \frac92\right)\left(a - b - \frac72\right) + \frac{63}{4} - 1 = 0.$ Multiply everything by $2\cdot 2 = 4$ : $(2a + 2b + 9)(2a - 2b - 7) = -59.$ Since both factors are integers and $59$ is prime, we have only four factorizations: $59 = \pm 59\cdot \mp 1$ or $\pm 1\cdot \mp 59$ . But $2a + 2b + 9$ is certainly greater than 9; therefore, only one possibility remains: $\begin{cases} 2a + 2b + 9 = 59 \\ 2a - 2b - 7 = -1 \end{cases}\ \ \ \therefore\ \ \ \begin{cases} a = 14 \\ b = 11 \end{cases}.$

Check the solution: $117_{14} = 14^2 + 14 + 7 = 217;\ \ \ 188_{11} = 11^2 + 8\cdot 11 + 8 = 217.$

Answer : We post $a = 14_{10} = \boxed{13}_{11}$ .

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Alternative solution . Suppose $a = b + d$ . We must solve $a^2 + a + 7 = (b + d)^2 + (b + d) + 7 = b^2 + (2d + 1)b + (d^2 + d + 7).$ Compare this to $188_b$ , accounting for the possible "carry" $c$ from the last to the middle digit: $\begin{cases} 2d + 1 + c = 8 \\ d^2 + d + 7 = bc + 8 \end{cases}$ From the first equation we see that $c$ is odd; in fact, the possibilities are limited to $(c,d) = (1,3);\ (3,2);\ (5,1)$ . Try solving the second equation for $b$ in each of these three cases:

• $3^2 + 3 + 7 = b + 8 \ \ \ \therefore\ \ \ 19 = b + 8\ \ \ \therefore\ \ \ b = 11$ ;

• $2^2 + 2 + 7 = 2b + 8$ has no solution, since left side is odd by right side is even.

• $1^2 + 1 + 7 = 3b + 8 \ \ \ \therefore\ \ \ 9 = 3b + 8\ \ \ \therefore\ \ \ b = \tfrac13$ is not an integer.

The only valid solution, then, is $b = 11$ and $d = 3$ , so that $a = \overline{1d}_b = \boxed{13}_{11}$ .

In this solution, I will be using the following statement repeatedly.

If a quadratic equation has integer solution, then its (quadratic) discriminant must be a perfect square.

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We want to find positive integers $(a,b)$ satisfying $a^2 + a + 7 = b^2 + 8b + 8$ . Rearranging this equation gives $b^2 (1) + b (8) + (-a^2 - a + 1) = 0$

The discriminant (in $b$ ) must be a perfect square. That is, there is an integer $c\geqslant 0$ satisfying

$8^2 - 4(1)(-a^2 - a + 1) = c^2 \quad \Rightarrow \quad 4a^2 + 4a + 60 = c^2$

Since LHS is divisible by 4, then so must RHS. Thus, $c$ must be even, $c = 2d$ ,

$4a^2 + 4a + 60 = (2d)^2 \quad \Rightarrow \quad a^2 + a + 15 = d^2 \quad \Rightarrow \quad a^2 (1) + a(1) + (-d^2+ 15) = 0$

Again, the discriminant must be a perfect square,

$1^2 - 4(1)(-d^2 + 15) = e^2 \quad \Rightarrow \quad (2d)^2 - e^2 = 59 \quad \Rightarrow \quad (2d+e)(2d-e) = 59$

Since 59 is prime, then

$\begin{cases} 2d + e = 59 \\ 2d - e = 1 \end{cases} \quad \Rightarrow \quad (d,e) = (15,29) \quad \Rightarrow \quad c = 30 \quad \Rightarrow \quad a = 14 \quad \Rightarrow \quad b = 11 \quad \Rightarrow \quad a = 11 + 3 = \boxed{13}_b$

Mathematica

For[x=4,x<300,x++, For[a=2,a<x,a++, For[b=2,b<x,b++, If[IntegerDigits[x,a]=={1,1,7}&&IntegerDigits[x,b]=={1,8,8}, Print[x, " ", "a=", a, " ", "b=", b]]]]]

217 a=14 b=11