Do there exist number bases $a$ and $b$ such that $111_a = 1111_b\ \ ?$

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I find it easier to just show that the cubic discriminant of the polynomial (in $b$ ) is always negative.

Pi Han Goh
- 3 years, 8 months ago

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Perhaps you should post a solution? I, for one, am interested to see it.

For every value of $b \geq 2$ there exists a positive real number $a = \frac{\sqrt{4(b^3 + b^2 + b) + 1} - 1}2$ such that $a^2 + a + 1 = b^3 + b^2 + b + 1$ ; the only problem is that it never is an integer.

Conversely, given $a \geq 2$ , define $q = a^2 + a + \frac7{27}$ , $D = q^2 + \frac{32}{729}$ , and $b = \sqrt[3]{\frac q2 + \frac12\sqrt D} + \sqrt[3]{\frac q2 - \frac12\sqrt D} - \frac 13,$ we have again a positive solution for the equation, but it is never an integer.

Arjen Vreugdenhil
- 3 years, 8 months ago

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Posted.

Ah Cardano's . I'm not a huge fan of it, but I appreciate this approach.

Pi Han Goh
- 3 years, 8 months ago

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@Pi Han Goh – I only use it to try and convince you that the equation always has at least one real solution, contrary to what you claim.

Arjen Vreugdenhil
- 3 years, 8 months ago

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@Arjen Vreugdenhil – Whoops, I didn't realize my error until you literally spelled it out for me.

Pi Han Goh
- 3 years, 8 months ago

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Disclaimer:
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As pointed out by Arjen, my Method 1 is completely wrong! But my method 2 still works! Yayyyy!!

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Method 1:
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The equation in question is equivalent to $a^2 + a + 1 = b^3 + b^2 + b + 1 \Leftrightarrow b^3 + b^2 + b + (-a^2 - a) = 0$ .

Suppose there exist number bases $a$ and $b$ , then the cubic equation above (in $b$ ) must have a non-negative discriminant . That is

$1 - 4 - 4 (-a^2 - a) - 27(-a^2 - a)^2 + 18(-a^2 - a) \geq 0 .$

Simplifying this inequality gives $27 a^4 + 54 a^3 + 41 a^2 + 14 a + 3 \leq 0$ . But since every single term in LHS is a positive number, the inequality cannot be fulfilled. A contradiction!

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Method 2:
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The equation in question is equivalent to $a^2 + a + 1 = b^3 + b^2 + b + 1 \Leftrightarrow a^2 + a + (-b^3 - b^2 - b) = 0$ .

Suppose there exist number bases $a$ and $b$ , then the discriminant above (in $a$ ) must be a perfect square. That is, for some integer $c$ ,

$1^2 - 4(-b^3 - b^2 - b) = c^2 \quad \Leftrightarrow\quad 1-c^2 = 4(-b^3 - b^2 - b).$

If $c$ is even, then LHS is odd and RHS is even, which is absurd. Thus $c$ must be odd. Let $c = 2d+1$ for some integer $d$ , then the equation simplifies to

$1 - (2d + 1)^2 = 4(-b^3 - b^2 - b) \quad \Leftrightarrow\quad b^3 + b^2 + b + 1 = d^2 + d + 1 ,$

which is where we started with (just replacing $a$ with $d$ ). So, there's no solution by infinite descent .

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Solution 1 won't work: A cubic equation has $\begin{cases} \text{one real solution} & \text{if}\ D < 0 \\ \text{three real solutions} & \text{if}\ D > 0 \end{cases}$

(Proof that there is always one real solution: let $f(x) = ax^3 + O(x^2)$ with $a > 0$ . Then there exist $x_- < x_+$ such that $f(x_-) < 0$ and $f(x_+) > 0$ . Since $f$ is continuous on $[x_-,x_+]$ , the middle-value theorem says there exists $x_- < x_0 < x_+$ such that $f(x_0) = 0$ . The argument for $a < 0$ is similar.)

Arjen Vreugdenhil
- 3 years, 8 months ago

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Ah, sorry. For some reason, I was thinking of it from a "quadratic discriminant's point of view".

I won't remove "Method 1", but I'll leave a disclaimer for others to read.

Pi Han Goh
- 3 years, 8 months ago

Solution 2: Nice use of infinite descent!

Arjen Vreugdenhil
- 3 years, 8 months ago

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If there were such a pair $(a,b)$ , it would satisfy the equation $a^2 + a + 1 = b^3 + b^2 + b + 1.$ Subtracting one and factoring gives $(a+1)a = (b^2 + b + 1)b.$ Let $d$ be the greatest common divisor of $a$ and $b$ ; write $a = a'd$ and $b = b'd$ . Then $(da'+1)a' = (eb' + 1)b'\ \ \ \ \ \ \ \ \text{with}\ \ \ \ e = d(b^2 + b).$ Since $da'+1$ is one more than a multiple of $a'$ , they are coprime; the same holds true for $e'b+1$ and $b'$ . Moreover, $a'$ and $b'$ are coprime. But then this equation can only be satisfied if $da'+1 = b',\ \ \ a' = eb' + 1.$ This would imply that $a' > b'$ and $b' > a'$ , which is impossible.