1111 = 111?

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.

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

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Method 1:

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:

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 .