Easy problem if d=gcd(a,b) a^2-ab+b^2|(ab)^2 iff a^2-ab+b^2|d^4 Now for number a a^2-ab+b^2<a^4 ut this holds for finite b

$ (a,b)=d, a=d.a_{1},b=d.b_{1}$ $ \implies$ $ (a_{1},b_{1})=1$ $ \implies$ $ d^{2}(a_{1}^{2}+b_{1}^{2}-a_{1}.b_{1})|d^{4}.a_{1}^{2}.b_{1}^{2}$ $ \implies$ $ ((a_{1}^{2}+b_{1}^{2}-a_{1}.b_{1}),a_{1}^{2}.b_{1}^{2})=1$ $ \implies$ $ (a_{1}^{2}+b_{1}^{2}-a_{1}.b_{1})|d^{2}$ $ \Box$

hossein11652 wrote: does there exist an infinite set S such that for every a, b in S we have a ² + b ² - ab | (ab) ² Easy problem but everyone slove problem: Let $ n$ positive integer.Prove or disprove that set $ S =\{ x_1,x_2,..,x_n\}$ such that $ gcd(x_i;x_j) = 1$ and $ x_i^2 + x_j^2 - x_ix_j|(x_ix_j)^2$

I think everyone who solved the first problem would have created problem two as well for I came up with the same problem. The solution is very simple. Claim: The set $ S$ can have at most one non-zero integer. Proof of Claim: Suppose not. Let $ x_1,x_2 \in S$ with $ x_1 \neq x_2$ and $ x_1,x_2 \neq 0$. $ gcd(x_1,x_1^2 + x_2^2 - x_1x_2) = gcd(x_1,x_2^2) = 1$ Similarly, $ gcd(x_2,x_1^2 + x_2^2 - x_1x_2) = gcd(x_2,x_1^2) = 1$ Hence, $ x_1^2 + x_2^2 - x_1x_2|(x_1x_2)^2 \Rightarrow x_1^2 + x_2^2 - x_1x_2|1$ We note that $ x_1$ and $ x_2$ should have the same sign. Otherwise, $ x_1^2 + x_2^2 - x_1x_2 \geq 3$ which is not possible. Therefore, $ 0 < x_1x_2 \leq x_1^2 + x_2^2 -x_1x_2$ and $ x_1^2 + x_2^2 - x_1x_2|1$. This further implies that $ 0 < x_1x_2 \leq x_1^2 + x_2^2 -x_1x_2 \leq 1\Rightarrow x_1x_2 = 1 \Rightarrow x_1 = x_2 = 1$ or $ x_1 = x_2 = -1$. This contradicts $ x_1 \neq x_2$. Hence, $ S$ has at most one non-zero integer. Therefore, $ S = \{0,a\}$ where $ a \neq 0$. Hence, for $ n>3$, there does not exist a set $ S$ with the required property.

Omid Hatami wrote: Now for number a a^2-ab+b^2<a^4 ut this holds for finite b Why does it hold for finite b?! Another thing:RHS isn't a^4,but d^4 Thanks.

No. Your "correction" ($d^4$ instead of $a^4$ in the RHS) would make that proof invalid. I agree Omid's proof is kind of terse, so I will make it clearer. Take $a,b \in S$, $a\neq 0$, $a\neq b$. Take $d=\gcd(a,b)$, $a=dx$, $b=dy$, so $\gcd(x,y)=1$. We then need $x^2 + y^2 - xy \mid d^2 x^2y^2$, and since $\gcd(x^2 + y^2 - xy, x^2y^2) = 1$, this forces $x^2 + y^2 - xy \mid d^2$, so $a^2 + b^2 - ab \mid d^4$. This alone does not yet directly lead to the conclusion it can only hold for finitely many $b$, since the value of $d$ may change with $b$. But we have $d\mid a$, so $d^4\leq a^4$. Having $a^2 + b^2 - ab \mid d^4$ forces $|a^2 + b^2 - ab| \leq d^4\leq a^4$. However, then $a_4 \geq |a^2 + b^2 - ab| = (a-b/2)^2 + 3b^2/4 \geq 3b^2/4$, so $b^2 \leq 4a^2/3$, and with $a$ fixed, it can only occur for finitely many $b$.

Thanks but i have problem in the last line.a isn't fixed so how can you suppose that it's fixed?

Why is $a$ not fixed? I took $a\in S$, $a\neq 0$. I am allowed to fix it, no?

Oh,I'm sorry...I had a mistake

Omid Hatami wrote: a^2-ab+b^2|(ab)^2 iff a^2-ab+b^2|d^4 Reason?