AIMO 2015 Q1

Note that since there is a $6$ in $146_{b}$ , $b \geq{7}$ .

$\Rightarrow 146_{b} \geq{7^{2}+4 \cdot 7+6}$ , $146_{b} \geq {83}$

Since there is a $3$ in $123_{a}$ , $a \geq{4}$ .

$\Rightarrow 123_{a} \geq{4^{2}+2 \cdot 4+3}$ , $123_{a} \geq {27}$

Clearly, $123_a, 146_b \geq 83$

Therefore, $146_{b} \geq{7^{2}+4 \cdot 7+6}$ , $146_{b} \geq {83}$

Since $123_{a} =146_{b}$ , $123_{a} \geq{83}$

Therefore, $a^2 + 2 \cdot a +3 \geq 83$ , and

$(a+10)(a-8) \geq 0$

$a \ge 0 \Rightarrow a \geq 8$

Therefore, there is a clearly a possible integer value for $a$ when $b=7$ . Namely, $a=8$ and $a + b = 15$ . Since this value is derived the absolute minimum value of $b$ and the minimum value of $a$ (which is greater than the absolute minimum of $a$ ) when $b$ is at its absolute minimum, $15$ is the absolute minimum of $a+b$ .

Note: (Not really important to the solution, rather, to its integrity) This solution may seem random, however, since it is known that there is a solution such that $a,b \in \mathbb{N}$ and we are trying to find the minimum values of $a+b$ , this algorithm of calculating a minimum and using it with the other equation to calculate a new, larger minimum could be used until integers were found that satisfies $a^2 + 2 \cdot a + 3 = b^2 +4 \cdot b +6$ . As an example, if the minimum of $123_a$ indicated that the minimum of $b$ , calculated using the quadratic formula, was not an integer, it could be rounded up to the next lowest integer and then this could be used to calculate a new minimum for $a$ , etc.

First, observe that b > 6 and therefore a > 7. Now it's obvious that

a^2 + 2a + 3 = b^2 + 4b + 6

a^2 + 2a + 3 - 2 = b^2 + 4b + 6 - 2

a^2 + 2a + 1 = b^2 + 4b + 4

(a + 1)^2 = (b + 2)^2

a = b + 1

b > 6, so the minimum of b is 7 and the minimum of a is 8.