29. $[\mathbf{1 7}]$ Let $a, b, c$ be positive integers. All the roots of each of the quadratics
$$
a x^{2}+b x+c, a x^{2}+b x-c, a x^{2}-b x+c, a x^{2}-b x-c
$$
are integers. Over all triples $(a, b, c)$, find the triple with the third smallest value of $a+b+c$.
Answer: $(1,10,24)$
The quadratic formula yields that the answers to these four quadratics are $\frac{ \pm b \pm \sqrt{b^{2} \pm 4 a c}}{2 a}$. Given that all eight of these expressions are integers, we can add or subtract appropriate pairs to get that $\frac{b}{a}$ and $\frac{\sqrt{b^{2} \pm 4 a c}}{a}$ are integers. Let $b^{\prime}=\frac{b}{a}$ and $c^{\prime}=\frac{4 c}{a}$. We can rewrite the expressions to get that $b^{\prime}$ and $\sqrt{b^{\prime 2} \pm c^{\prime}}$ are positive integers, which also tells us that $c^{\prime}$ is a positive integer. Let $b^{\prime 2}+c^{\prime}=n^{2}$, $b^{\prime 2}-c^{\prime}=m^{2}$.
Notice that $a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right)$, so to find the third smallest value of $a+b+c$, we first find small solutions to $\left(b^{\prime}, c^{\prime}\right)$. To do this, we find triples $\left(m, b^{\prime}, n\right)$ such that $m^{2}, b^{\prime 2}, n^{2}$ form an arithmetic sequence. Because odd squares are $1 \bmod 4$ and even squares are $0 \bmod 4$, if any of these three terms is odd, then all three terms must be odd. By dividing these terms by the largest possible power of 2 then applying the same logic, we can extend our result to conclude that $v_{2}(m)=v_{2}\left(b^{\prime}\right)=v_{2}(n)$. Thus, we only need to look at $\left(m, b^{\prime}, n\right)$ all odd, then multiply them by powers of 2 to get even solutions.
We then plug in $b^{\prime}=3,5,7,9$, and find that out of these options, only $\left(n, b^{\prime}, m\right)=(1,5,7)$ works, giving $\left(b^{\prime}, c^{\prime}\right)=(5,24), a+b+c=12 a$. Multiplying by 2 yields that $\left(n, b^{\prime}, m\right)=(2,10,14)$ also works, giving $\left(b^{\prime}, c^{\prime}\right)=(10,96), a+b+c=35 a$. For $11 \leq b \leq 17$, we can check that $m=b+2$ fails to give an integer $\mathrm{n}$. For $11 \leq b \leq 17, m \neq b+2, a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right) \geq a\left(1+11+\frac{15^{2}-11^{2}}{4}\right)=38 a$, the smallest possible value of which is greater than $12 a$ with $a=1,12 a$ with $a=2$, and $35 a$ with $a=1$. Thus, it cannot correspond to the solution with the third smallest $a+b+c$. For $b \geq 19$, $a+b+c=a\left(1+b^{\prime}+\frac{c^{\prime}}{4}\right) \geq a\left(1+19+\frac{21^{2}+19^{2}}{4}\right)=40 a$, which, similar as before, can't correspond to the solution with the third smallest $a+b+c$.
Thus the smallest solution is $\left(a, b^{\prime}, c^{\prime}\right)=(1,5,24),(a, b, c)=(1,5,6)$, the second smallest solution is $\left(a, b^{\prime}, c^{\prime}\right)=(2,5,24),(a, b, c)=(2,10,12)$, and the third smallest solution that the problem asks for is $\left(a, b^{\prime}, c^{\prime}\right)=(1,10,96),(a, b, c)=(1,10,24)$.