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# Vieta Root Jumping Help Needed

Hey everyone, I was working on writing a problem on the above topic, and needless to say I am not proficient enough in it to check my own work, and I'm also unsure as to how to prove that all possibilities have been accounted for.

Positive integers $$a,b,c\leq2016$$ satisfy the following equation:

$\frac{a^{2}+b^{2}+1}{abc-1}=1$

What is the largest possible value of $$b$$?

My Attempt at a Solution:

Lemma 1:

Suppose for a fixed $$c$$ that a pair of integers $$a,b$$ satisfy the equation. Then $$b, d=\frac{b^{2}+2}{a}$$ is also a solution.

Proof:

We can rearrange our condition as $$a^{2}-abc+b^{2}+2=0$$. Let $$a=x$$.

Then $$x^{2}-bcx+b^{2}+2=0$$ has solutions $$a,d$$ for some value $$d$$.

We know by Vieta's Formula that the sum of the roots is $$a+d=bc$$, so $$d\in \mathbb{Z}$$

we also know that the by the product of the roots $$d=\frac{b^{2}+2}{a}$$.

Hence, $$b,d$$ is also a solution. $$\square$$

Lemma 2:

If $$a>b\geq3$$, then $$0<d< b$$

Proof:

$$d=\frac{b^{2}+2}{a}<b+\frac{2}{b}<b+1\leq a$$. Since $$d$$ is an integer, then it also must be less than or equal to $$b$$, since the next highest $$d$$ could be is $$b+1$$, which is outside of the bounds.

For $$d=b$$, we get that $$2=b^{2}(c-2)$$, and it is easy to see that $$b=d=1,c=4$$ is the only solution, but $$a>b\geq3$$, a contradiction.

$$d=\frac{b^{2}+2}{a}$$, and since $$b$$ and $$a$$ are positive, $$d$$ must also be positive, therefore $$d>0$$.

Hence $$0<d<b$$ for $$a>b\geq3$$. $$\square$$

Now, suppose $$a,b,c$$ satisfy the above equation. We only need to check for $$b=1,2$$, since if $$b\geq3$$, there exists a smaller solution.

• Case 1: $$b=1 \rightarrow a^{2}+3=ac \rightarrow a+\frac{3}{a}=c$$. Since $$a,c \in \mathbb{N}$$, we know that $$a=1,3$$, are the only possibilities, therefore $$c=4$$.

• Case 2: $$b=2 \rightarrow a+\frac{6}{a}=2c$$. For $$a+\frac{6}{a}$$ to be an integer, $$a=1,2,3,6$$, so $$2c=7,5$$, both of which are odd, therefore $$c$$ is not an integer and we have no solutions.

This means that $$c=4$$ will be the only value of $$c$$ which produces integer solutions for $$a,b$$*

Now to find the greatest possible value of $$b\leq2016$$, we simply have to jump our way up the roots of this one and only chain, this time interchanging $$a$$ and $$b$$ in our formulas in order to increase the sequence rather than decrease it.

This means the next solution $$a_{1}=\frac{a^{2}+2}{b}$$, then $$b_{1}=\frac{a_{1}^{2}+2}{b}$$, then $$a_{2}=\frac{a_{1}^{2}+2}{b_{1}}$$, and so on until either $$a$$ or $$b$$ surpasses $$2016$$.

$$(a_{1},b_{0})=(3,1) \\ (a_{1},b_{1})=(3,11) \\(a_{2},b_{1})=(41,11) \\(a_{2},b_{2})=(41,153) \\(a_{3},b_{2})=(571,153) \\(a_{3},b_{3})=(571,2131)$$

$$2131>2016$$, so our maximum value before $$2016$$ is $$\boxed{571}$$.

*How do we know that there isn't some chain or even outliers that don't jump down to our lemma? Like perhaps $$a=11,b=34,c=137$$? (I know this isn't a solution but just random numbers).

Is that because I showed that if $$a>b\geq3$$ has a solution in $$c$$, that for that same $$c$$ we can reduce both $$a$$ and $$b$$ below $$3$$?

Thanks in advance for the help!

Note by Brandon Monsen
6 days, 7 hours ago

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Yeah I'm pretty sure this proof works. You just need to tidy up a few bits and pieces, for example the statement of lemma one should read:

Suppose for a fixed $$c$$ a pair of integers $$a,b$$ satisfy the condition. Then $$a,b=(d=\frac{b^2+2}{a}),b$$ also satisfy the condition.

The first time I read that sentence it took me about ten seconds to realise it wasn't $$b,d$$ both equal to $$\frac{b^2+2}{a}$$ · 3 days, 3 hours ago

Okay, I'll leave this up for another day or so if someone else wants to chime in, and then I'll take it down before I post the actual problem. Thanks :) · 2 days, 16 hours ago

If you look at vieta root jumping, it is stated that

Using Vieta’s formula, we can display a second solution to this equation. The next step is to show that the new solution is valid and smaller than the previous one. Then by the argument of infinite descent or by assuming the minimality of the first solution, we get a contradiction.

The special part is defining what makes the solution "smaller". Often, it is sufficient to just look at $$\max(a,b)$$ to order the elements in the chain (but finding a partial ordering for equations with 3 variables could be harder). Staff · 1 day, 19 hours ago

It was the triple variable part that was getting to me. I started by assuming that there was a solution for a fixed c, which I'm assuming is okay?

I then showed for that same C that unless both variables are 3 or more that there will be another solution for that same value of c. Would it work to say that I would then only have to check for values of c at b=1,2? · 1 day, 17 hours ago

Your solution is correct. By fixing $$c$$ in lemma's 1 and 2, you avoid having to deal with a triple chain (and it makes sense, because the equation isn't cyclic/symmetric). Essentially you have shown (perhaps needing to add a bit more clarification for those not familiar with how this works), that if we have a solution $$(A, B, C)$$, then there is a corresponding solution $$(a,b,c)$$ with $$c = C$$ and $$b \leq 2$$. Staff · 1 day, 16 hours ago

Ok thanks! Im gonna delete this thread later on tonight and probably post the problem then. · 1 day, 15 hours ago