Six integers $a,b,c,d,e,f$ satisfy: $\begin{cases} ace+3ebd-3bcf+3adf=5 \\ bce+acf-ade+3bdf=2 \end{cases}$ Find all possible values of $abcde$ D. Bazyleu
Problem
Source: Belarus TST 2024
Tags: algebra, system of equations, number theory
26.07.2024 08:55
How are you supposed to solve this problem? I'm guessing that the answer is 0, but I have no idea how to prove that.
26.07.2024 10:57
chenghaohu wrote: How are you supposed to solve this problem? I'm guessing that the answer is 0, but I have no idea how to prove that. Yes, the answer is indeed zero.
26.07.2024 21:51
Unfortunately I still can't solve the problem. What's the correct set of manipulations to do?
28.07.2024 02:38
nAalniaOMliO wrote: After you have done hint 1, you should
How are you supposed to factorize this expression? I tried for a while and got nowhere. Can you PM me the solution? I would like to know what trick finishes the problem.
28.07.2024 12:45
I am really sorry.
28.07.2024 13:19
$(ace+3ebd-3bcf+3adf)^2+3(bce+acf-ade+3bdf)^2=37=(a^2+3b^2)(c^2+3d^2)(e^2+3f^2)=1*1*37$ Solving it over integers, we can find, that $(|a|,|b|,|c|,|d|,|e|,|f|)$ is premutation of $(5,2,1,0,1,0)$ , so $abcde = 0$
28.07.2024 13:35
What a trick! How many participants are solved this problem? I guess author started from the following claim: product of numbers in form $x^2+Ay^2$ can be represented in the same form and then created the problem. Is it true?
28.07.2024 14:29
Out of 18 participants of TST, 10 got 7/7, 4 got 5/7, 1 got 4/7, 1 got 2/7, 2 got 1/7 You should understand that it is not the only possible solution to this problem. For example, half of the IMO team had the following solution: Note that $(a+b\sqrt{-3})(c-d\sqrt{-3})(e+f\sqrt{-3})=ace+3ebd-3bcf+3adf+(bce+acf-ade+3bdf)\sqrt{-3}=5+2\sqrt{-3}$, now you write the norms and you are done. The majority of other solutions were cases-based, where if you spend some time you eventually will get a contradiction to every case where variables are non-zero. If you open the Belarus TST 2024 folder, you will find that this problem is actually the first out of four on its test So all students started solving it immediately in the beginning of the contest and had all 5 hours to solve it.