Peter wrote: If $ a, b, c$ are positive integers such that \[ 0 < a^{2}+b^{2}-abc \le c, \] show that $ a^{2}+b^{2}-abc$ is a perfect square. http://www.mathoe.com/dispbbs.asp?boardid=117&id=10850&star=1#15926

Can you translate that solution please?

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=605111911&t=6233

I use a modification of the Vieta approach for A3 (probably featuring in some of those links). Assume a solution exists and let $ a^{2}+b^{2}-abc=q\Leftrightarrow a^{2}+b^{2}-abc-q=0$. By symmetry, we may assume $ a\geq b$ and I modify the notation, letting $ a_{0}=a, a_{1}=b$, we get (setting n=0) \[ a_{n}^{2}+a_{n+1}^{2}-a_{n}a_{n+1}c-q=0\] Considering as a quadratic in $ a_{n}$, and letting $ a_{n+2}$ be the other root, we obtain the relations:\[ a_{n+2}=a_{n+1}q-a_{n}=\frac{a_{n+1}^{2}-q}{a_{n}}\] It is clear that $ a_{n+2}<a_{n+1}$ (since q>0) and that it is an integer. We can continue doing this indefinitely, making the above equation meaningful for all n and determining a decreasing sequence $ \{a_{i}\}$ of integers. Suppose this sequence never reaches 0. Then there exists $ k$ such that $ a_{k}>0>a_{k+1}$. Considering the above quadratic with $ n=k-1$, we get (taking the lesser root) \[ q=a_{k}^{2}+a_{k+1}^{2}-a_{k+1}a_{k}c > a_{k}c\geq c\] This contradicts the given fact that $ q\leq c$, so we must at some point reach zero, say, at $ n=l, a_{l+2}=0$. Here we get $ a_{l}a_{l+2}=a_{l+1}^{2}-q=0$, whence $ q=a_{l+1}^{2}$, so $ q$ is a perfect square, as required.

Let me rewrite that a bit easier. Assume there exists $ a,b,c,q > 0$ with $ c\ge q$ such that $ a^2 + b^2 - abc = q$ but $ q$ is not a perfect square (else there is nothing to prove). Take them such that $ a + b$ is minimal and assume wlog $ a\ge b$. Solving this equation for $ a$ we get two roots: $ a,a' = \frac {bc\pm \sqrt {b^2c^2 - 4(b^2 - q)}}2$, and since one is integer, so is the other. Since $ bc(a - bc) + b^2 - q\ge bc + b^2 - q\ge b^2 + (b - 1)c\ge0 = a(a - bc) + b^2 - q$ we need $ bc\ge a$ and thus $ b^2 > q$ (since $ q\not = b^2$ by assumption) and thus $ a,a' > 0$. Now $ a' + b\ge a + b$ by minimality of $ a + b$, while $ aa' = b^2 - q\le b^2$ by de Vieta, so $ a'\le b$. As we had $ a'\ge a\ge b$, we have $ a' = b = a\Rightarrow q = 0$, contradiction.

How do you reach the conclusion $ bc(a-bc)\geq bc$?

Beautiful solution in http://www.mathoe.com/dispbbs.asp?boardid=117&id=10850&star=1#15926 translated and repackaged:- Let $ a^2 + b^2 - abc =: d$ and $ \mathbb{Z^+} \ni d \le c$, so $ a^2 - abd + b^2-d = 0$. Let $ (A,B) \in \mathbb{Z^+} \times \mathbb{Z^+}\math$ be a solution to \[ X^2 - cXY + Y^2-d = 0 \cdots \cdots \cdots \cdots \cdots [1] \] with $ B$ minimum positive (using Well-Ordering Principle). By symmetry, $ (B,A)$ is also a solution, so $ B \le A$ by minimality of $ B$. Let $ M$ be another solution besides $ A$ to the quadratic in one variable \[ x^2 - cBx + B^2-d = 0 \cdots \cdots \cdots \cdots \cdots [2] \] Then from the Vieta relations $ M+A=cB$ and $ MA=B^2-d$, $ M=cB-A \in \mathbb{Z}$, $ M = B^2/A - d/A < B^2/A \le B$, since $ (B,M)$ also satisfies [1]. By the mimimality of $ B$, we can only have $ M \le 0$. And since $ (1+A)(1+M) = 1+A+M+AM = 1 + cB + B^2 - d > cB - d \ge c - d \ge 0$, we have $ 1+M>0$ i.e. $ M = 0$. Substituting into [2], we obtain $ d=B^2$. Thus $ a^2 + b^2 - abc$ is a square number. $ \blacksquare$

You neglected the case $ M=B$, and also there is a rather confusing typo in the first line (replace d with c)

me@home wrote: How do you reach the conclusion $ bc(a - bc)\geq bc$? You are right that it's not really clear. Let me explain: assume for the sake of contradiction that $ a > bc$ (thus $ a - bc\ge 1$). Then we have Peter wrote: $ bc(a - bc) + b^2 - q\ge bc + b^2 - q\ge b^2 + (b - 1)c\ge0 = a(a - bc) + b^2 - q$ and hence $ bc\ge a$, contradiction. So the only possibility is $ bc\ge a$.

me@home wrote: You neglected the case $ M = B$, and also there is a rather confusing typo in the first line (replace d with c) Yes, "$ a^2 - abd + b^2-d = 0$" should read "$ a^2 - abc + b^2-d = 0$". But wouldn't the chain $ M = B^2/A - d/A < B^2/A \le B$ imply that $ M < B$?

Let $ k=a^2+b^2-abc$ Let $ a_0,b_0,c_0$ be other 3 numbers such that our condition and $ a_0>b_0$ ,$ a_0+b_0$ is the least value so $ a+a_0=bc,aa_0=b^2-k$ if $ a_0<0$ thus we have $ k-b^2\ge bc$ which is contraction because $ k<c$ if $ a_0 >0$ thus $ a>a_0>b$ thus $ (a-1)(a_0-1)=b^2-k-bc+1\ge b^2$which is contraction again so $ a=0$ thus $ k=b^2$ is a perfect square

Why this problem is presented here in such incomplete form? More generally statement is true. Namely, if $a$, $b$, $c$ are positive integers such that \[ |a^2+b^2-abc-2|<c, \] then $a^2+b^2-abc$ is a perfect square.

nnosipov wrote: Why this problem is presented here in such incomplete form? More generally statement is true. Namely, if $a$, $b$, $c$ are positive integers such that \[ |a^2+b^2-abc-2|<c, \] then $a^2+b^2-abc$ is a perfect square. Why \[ |a^2+b^2-abc-2|<c, \] then $a^2+b^2-abc$ is a perfect square

Allnames wrote: Let $ k=a^2+b^2-abc$ Let $ a_0,b_0,c_0$ be other 3 numbers such that our condition and $ a_0>b_0$ ,$ a_0+b_0$ is the least value so $ a+a_0=bc,aa_0=b^2-k$ if $ a_0<0$ thus we have $ k-b^2\ge bc$ which is contraction because $ k<c$ if $ a_0 >0$ thus $ a>a_0>b$ thus $ (a-1)(a_0-1)=b^2-k-bc+1\ge b^2$which is contraction again so $ a=0$ thus $ k=b^2$ is a perfect square Why $ k-b^2\ge bc$