Another Base a and b

This involved some sneaky steps so I've tried to explain as much as possible as I go below - you could write this solution in a few lines but it might not be the easiest to understand! Let me know if anything's not clear.

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Writing the numbers out in full, we have $a^2+4a+8=b^2+9b+8$

We want to solve this in positive integers, with $a>8$ and $b>9$ . Since there's only one equation in two unknowns, we'll have to use properties of integers to solve it; we want to limit the number of possible solutions as much as possible. (This more or less rules out treating this as a quadratic, for example.)

This idea motivates the following manipulation: $\begin{aligned} a^2+4a+8&=b^2+9b+8 & \\ a^2+4a+4&=b^2+9b+4 &\text{recognising a square on the left} \\ (a+2)^2 &=b^2+9b+4 & \\ 4(a+2)^2 &= 4b^2+36b+16 &\text{wanting a square on the right} \\ (2a+4)^2 &= (2b+9)^2-81+16 & \\ (2a+4)^2 &= (2b+9)^2-65 & \\ (2b+9)^2-(2a+4)^2 &=65 & \end{aligned}$

So we have two integers, $x=2b+9$ and $y=2a+4$ , with a difference of squares equal to $65$ ; ie $(x+y)(x-y)=65$ .

This does the trick - $65=5\cdot 13$ only factorises in a few ways, so we have the reduction to case analysis.

Further, note that since $a$ and $b$ are positive, and the two factors of $65$ have to have the same sign, we must have $x+y>x-y>0$ ; so there are only two cases to consider.

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Case 1: $x+y=13$ , $x-y=5$

Solving these leads to $x=9$ , $y=4$ , from which we get $a=0$ and $b=2$ ; but this solution isn't valid as $a$ is not positive.

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Case 2: $x+y=65$ , $x-y=1$

Solving these leads to $x=33$ , $y=32$ , from which we get $a=14$ and $b=12$ , which works. To check, we get $14^2+4\times 14 + 8 = 12^2+9\times 12+8=\boxed{260}$ which is the required answer.

Let the given number be $N$ . It is given that

$\begin{aligned} 148_a & = 198_b \\ a^2 + 4a + 8 & = b^2 + 9b + 8 \\ a^2 + 4a & = b^2 + 9b \\ a(a+4) & = b(b+9) \end{aligned}$

Note that the right-hand side $b(b+9)$ is always even (when $b$ is even or odd). Therefore the left-hand side $a(a+4)$ is always even too. But the left-hand side is only even when $a$ is even. When $a$ is even and $a+4$ is also even, this means that the left-hand side is always divisible by $4$ therefore the right-hand side $b(b+9)$ is also always divisible by $4$ . Since the number $198_b$ has $9$ as a digit, $b \ge 10$ . Then we have:

$\begin{aligned} a(a+9) & = b(b+9) \\ a^2 + 4a - b(b+9) & = 0 \\ \implies a & = \sqrt{4 + \blue{b(b+9)}} - 2 & \small \blue{\text{where }4 \mid b(b+9) \text{ and }b \ge 10} \end{aligned}$

The smallest $b \ge 10$ , where $4 \mid b(b+9)$ , is $11$ . Then $a = \sqrt{224}-2$ not an integer. The next $b$ is $12$ ; and $a = \sqrt{256}-2 = 14$ . Then $N = 14^2 + 4(14)+8 = 12^2 + 9(12)+8 = \boxed{260}$ .

Note: My method does not prove that the solution is unique.

148 in base a = 198 in base b
a² + 4a + 8 = b² + 9b + 8
a(a + 4) = b(b + 9)

Since RHS is a multiplication between an odd and an even number, a must be a positive even number. Also, since the difference between the numbers in RHS is 9, either both are multiple of 3 or both aren't. Supposing b isn't a multiple of 3, then neither a nor a+4 can be one, too. This would only be possible when their mean, (a + (a + 4)) / 2 = a + 2 is a multiple of 3 or a = 3x – 2 and a + 4 = 3x + 2 for a LHS of (3x – 2)(3x + 2)
= 9x² – 4
= one less than a multiple of 3.

Meanwhile, RHS would be (3y ± 1)(3z ± 1)
= 9yz ± 3(y + z) + 1
= one more than a multiple of 3
≠ the result in LHS
==> The supposed RHS with 2 non-multiples of 3 is impossible
[ Also can be done as (3y + 1)(3y + 10) and/or (3y – 1)(3y + 8) for a result of 9y² + 33y + 10 and 9y² + 21y – 8 respectively, which can be reduced to { 3(3y² + 11y + 3) + 1 } for the former and { 3(3y² + 7y – 3) + 1 } for the latter ].

Therefore, other than the fact that the even number between b and b+9 is a multiple of 4 (from the double evens of a and a+4 of LHS), b and b+9 of RHS also influenced the LHS in return by them being two multiples of 3 making one of a or a+4 to be a multiple of 3² = 9 (can't have the two in LHS to carry the factors of 3 separately since their difference is 4).

Then the least the LHS pair could be is 14 × 18 and the next should be 18 × 22. The first have a solution of 14 × 18 = 12 × 21 but none found for the second one (18 = 2 × 3² & 22 = 2 × 11, taking out the 3s we need the rest of 2, 2 and 11 to be divided into two groups in which their products differ by 3 but the smallest difference is 11 – 2² = 7 ≠ 3).

Now having a = 14 and b = 12, in base 10 the number equals
= 1 × 14² + 4 × 14¹ + 8 × 14⁰
= 196 + 56 + 8
= 260

Or = 1 × 12² + 9 × 12¹ + 8 × 12⁰
= 144 + 108 + 8
= 260

From given data, we can form 2 equations, a²+4a+8=b²+9b+8 => a+2= $\sqrt {b²+9b+4}$ . And a>b>9 [since, 9 is representable in base b, b must be bigger than 9, and 148<198, thus a must be bigger than b]. But since b>9, b²+6b+9<b²+9b+4<b²+10b+25, but the middle term is bounded by 2 perfect squares and it must be a perfect square in itself, thus only possibility is b²+9b+4=b²+8b+16 => b=12. Putting b=12 in 1st equation we have a=14 and the original number base 10 to be 260