What is the minimum value of a+b+c+d if a^2-b^2+cd^2=2022?

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Discussion Overview

The discussion revolves around finding the minimum value of the expression \(a+b+c+d\) given the equation \(a^2-b^2+cd^2=2022\), where \(a\), \(b\), \(c\), and \(d\) are non-negative integers. The scope includes mathematical reasoning and problem-solving related to this equation.

Discussion Character

  • Mathematical reasoning, Homework-related, Exploratory

Main Points Raised

  • One participant suggests \(a=1, b=2, c=1, d=45\) resulting in \(a+b+c+d=49\), but later acknowledges this guess is incorrect.
  • Another participant proposes \(a=17, b=1, c=6, d=17\) leading to \(a+b+c+d=41\) as their current best solution.
  • A different participant offers \(a=0, b=1, c=7, d=17\) which gives \(a+b+c+d=25\) as a potential minimum.
  • One participant expresses regret for not noticing a previous answer and mentions a wrong answer of 27 that they cannot delete.

Areas of Agreement / Disagreement

Participants do not appear to agree on a single minimum value, as multiple solutions have been proposed with varying results. The discussion remains unresolved regarding the optimal values of \(a\), \(b\), \(c\), and \(d\).

Contextual Notes

Some solutions may depend on specific assumptions about the values of \(a\), \(b\), \(c\), and \(d\), and there may be additional constraints not fully explored in the discussion.

anemone
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Here is this week's POTW:

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If ##a,\,b,\,c## and ##d## are non-negative integers and ##a^2-b^2+cd^2=2022##, find the minimum value of ##a+b+c+d##.

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Thanks for the interesting problem. I guess
a=1,b=2,c=1,d=45;\ a+b+c+d=49
I would like to know the right answer.
 
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anuttarasammyak said:
Thanks for the interesting problem. I guess
a=1,b=2,c=1,d=45;\ a+b+c+d=49
I would like to know the right answer.
I am sorry. Your guess is incorrect.

I will wait a bit longer before posting the answer to this POTW, just in case there are others who would like to try it out.
 
The best I can do for now is:$$a = 17, b = 1, c = 6, d = 17; \ a + b + c + d = 41$$
 
Last edited:
Well, I feel a bit guilty answering as it’s been an (extremely) long time since I was at school! However:
##a=0, b=1, c=7, d=17##
##a+b+c+d = 25##
 
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Opps! I didn't see @Steve4Physics 's answer above soon enough. Why can't I delete my wrong answer (27)?
 
Last edited:

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