Does Diophantus help?
Consider positive integers a , b , c , and d .
Is it possible that ∣ a c + b d ∣ = ∣ a d − b c ∣ ?
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3 solutions
A possibe solution:
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ a = 4 b = 2 c = 1 d = 3
Yes , for example ( a , b , c , d ) = ( 6 , 2 , 3 , 6 ) .
(And, no, I don't believe Diophantus helps at all... :-/ )
How did you find this? Are there infinitely many solutions?
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Yes, there are! Once you have found one solution, you can multiply all the numbers by any integer and that will also be a solution.
I would also be interested to know if there is a more direct/elegant method for solving. I must admit I resorted to trial and error.
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Let me rephrase my question: Are there infinitely many quintuplets of COPRIME positive integers (a,b,c,d) that satisfy this criteria?
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@Pi Han Goh – Ah yes, I believe there are, based on HM's solution above... Since there are infinitely many combinations of a and b you can choose that lead to coprime positive integers (a,b,c,d) that satisfy this criteria.
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@Geoff Pilling – Then, how do you know that there are infinitely many solutions to gcd(a,b,a-b,a+b) = 1? I don't think that's obvious.
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@Pi Han Goh – Hmmmmm... Lemme think about that one...
From this, I assume we are not going to make the assumption that a , b , c , d are distinct positive integers. Is that what you had in mind?
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That would be an interesting follow up question... Although I see that Áron has found a solution to the d i s t i n c t case above.
Without loss of generality, let a c + b d = a d − b c .
Then ( a + b ) c = ( a − b ) d .
For any positive integers a > b , if c = a − b and d = a + b , the equality holds. □