Problem 486

Suppose that $r = \sqrt{\frac{\sqrt{53}}{2} + \frac{3}{2}}$. Thus, $r^{2} = \frac{\sqrt{53}}{2} + \frac{3}{2}$ and so $2r^{2} = \sqrt{53} + 3$ or $2r^{2} - 3 = \sqrt{53}$. Squaring both sides again, we obtain $(2r^{2} - 3)^{2} = 53$ or $4r^{4} - 12r^{2} + 9 = 53$ which gives $4r^{4} - 12r^{2} - 44 = 0$ or $r^{4} - 3r^{2} - 11 = 0$ or $r^{4} = 3r^{2} + 11$. Suppose next that $r^{100} = 2r^{98} + 14r^{96} + 11r^{94} - r^{50} + ar^{46} + br^{44} + cr^{40}$ for some positive integers $a, b, c$. Since $r \neq 0$, we can divide by $r^{40}$ to obtain $r^{60} = 2r^{58} + 14r^{56} + 11r^{54} - r^{10} + ar^{6} + br^{4} + c$. Now using the relationship $r^{4} = 3r^{2} + 11$, we can see that $r^{60} - 2r^{58} - 14r^{56} - 11r^{54} = r^{54}(r^{6} - 2r^{4} - 14r^{2} - 11) = r^{54}(r^{2}(3r^{2} + 11) - 2r^{4} - 14r^{2} - 11) = r^{54}(3r^{4} + 11r^{2} - 2r^{4} - 14r^{2} - 11) = r^{54}(r^{4} - 3r^{2} - 11) = r^{54}(0) = 0$. Therefore, the equation is equivalent to the much simpler equation $r^{10} = ar^{6} + br^{4} + c$. Next, we express $r^{10}$ and $r^{6}$ as combinations of $r^{2}$ and constant terms. (To do this, we will need to express $r^{8}$ in this way too.) $r^{6} = r^{2}r^{4} = r^{2}(3r^{2} + 11) = 3r^{4} + 11r^{2} = 3(3r^{2} + 11) + 11r^{2} = 20r^{2} + 33$, $r^{8} = r^{2}r^{6} = r^{2}(20r^{2} + 33) = 20r^{4} + 33r^{2} = 20(3r^{2} + 11) + 33r^{2} = 93r^{2} + 220$, $r^{10} = r^{2}r^{8} = r^{2}(93r^{2} + 220) = 93r^{4} + 220r^{2} = 93(3r^{2} + 11) + 220r^{2} = 499r^{2} + 1023$. Therefore, the equation $r^{10} = ar^{6} + br^{4} + c$ is equivalent to $499r^{2} + 1023 = a(20r^{2} + 33) + b(3r^{2} + 11) + c$. Rearranging, we obtain $0 = r^{2}(20a + 3b - 499) + (33a + 11b + c - 1023)$. Therefore, if $20a + 3b = 499$ and $33a + 11b + c = 1023$, then the equation is satisfied. So the original problem is equivalent to finding positive integers $a, b, c$ with $20a + 3b = 499$ and $33a + 11b + c = 1023$. We proceed by finding pairs $(a, b)$ of positive integers that satisfy $20a + 3b = 499$ and then checking to see if the value of $c = 1023 - 33a - 11b$ is positive. Since we need to find one triple $(a, b, c)$ of positive integers, we do not have to worry greatly about justifying that we have all solutions at any step. Since $20a$ has a ones digit of 0 and $20a + 3b = 499$, then the ones digit of $3b$ must be 9, which means that the ones digit of $b$ must be 3. If $b = 3$, we obtain $20a = 499 - 3b = 490$ and so $a$ is not an integer. If $b = 13$, we obtain $20a = 499 - 3b = 460$ and so $a = 23$. Note that, from $(a, b) = (23, 13)$, we can obtain additional solutions by noticing that $20(3) = 3(20)$ and so if we decrease $a$ by 3 and increase $b$ by 20, the sum $20a + 3b$ does not change. However, it turns out that if $(a, b) = (23, 13)$, then $c = 1023 - 33(23) - 11(13) = 121$. Since we are only looking for a unique triple $(a, b, c)$, then $(a, b, c) = (23, 13, 121)$. Finally, $a^{2} + b^{2} + c^{2} = 23^{2} + 13^{2} + 121^{2} = 15339$.