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

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The problem presented involves finding the minimum value of a+b+c+d given the equation a^2-b^2+cd^2=2022, with a, b, c, and d as non-negative integers. Initial guesses included a=1, b=2, c=1, d=45, resulting in a sum of 49, which was incorrect. A better solution was proposed with a=17, b=1, c=6, d=17, yielding a sum of 41. Ultimately, the lowest verified sum was a=0, b=1, c=7, d=17, resulting in a total of 25. The discussion highlights the collaborative effort to solve the problem and correct previous mistakes.
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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.
 
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:$$a = 17, b = 1, c = 6, d = 17; \ a + b + c + d = 41$$
 
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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)?
 
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