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For example, suppose we have two numbers $x, y \in$ N with their corresponding bases $a$ and $b$

$ x_{a} = y_{b} $

How do you determine $a$ and $b$ such that the equation is satisfied?

I tried interpreting the problem as a equation, such that each digit in my numbers x and y are a coefficient. For example, I have the numbers $241$ and $27$, and my idea was that I try to solve the equation $2a^2+4a+1=2b+7$. Unfortunately, I can't seem to get past that.

adamt
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  • Hi and welcome to MSE. What have you tried? – Menezio May 10 '20 at 18:58
  • Hey, I tried interpreting the problem as a equation, such that each digit in my numbers $x$ and $y$ are a coefficient. For example, I have the numbers $241$ and $27$, and my idea was that I try to solve an equation $2 * a^2 + 4 * a^1 + 1 * a^0 = 2b^1 + 7b^0$. Unfortunately, I can't seem to get past that. – adamt May 10 '20 at 19:03
  • Adamt, add what you wrote in the comment into the post. – PinkyWay May 10 '20 at 19:14

1 Answers1

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For small numbers you can often make progress with divisibility. First note that $a^0=b^0=1$, so in your example we need $$2a^2+4a+1=2b+7\\2a^2+4a=2b+6\\a^2+2a=b+3\\b=a^2+2a-3$$ and you can choose any $a$ you want greater than $4$ and compute $b$. This will duly make $b \gt 7$. When the numbers have three digits you will get quadratic equations, which allow you to require that the square root be integral. Similarly here, if we wanted to start from $b$ we would find $$(a+1)^2=b+4\\a=\sqrt{b+4}-1$$ so we must choose $b$ to be four less that a square. We need $a\gt 4$, so $b \ge 32$

Ross Millikan
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  • Hey Ross, thank you for your answer! Unfortunately, I don't yet fully understand why I should choose $a > 4$ in this example. Could you perhaps explain? – adamt May 10 '20 at 20:28
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    Because you have a $4$ in the base $a$ expansion, so $a$ must be at least $5$ to allow that. Similarly we must have $b \gt 7$, but given the limit on $a$ that is always satisfied. – Ross Millikan May 10 '20 at 20:41