Find all positive integers n such that 1 2 n − 1 1 9 and 7 5 n − 5 3 9 are both perfect squares.
Let N be the sum of all possible values of n . Find N
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Same Solution
How did 20 and 12 come and which are the five cases to be considered
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The five cases to consider are:
2 l + 5 k = 8 1 9 and 2 l − 5 k = 1
2 l + 5 k = 2 7 3 and 2 l − 5 k = 3
2 l + 5 k = 1 1 7 and 2 l − 5 k = 7
2 l + 5 k = 6 3 and 2 l − 5 k = 1 3
2 l + 5 k = 3 9 and 2 l − 5 k = 2 1 .
When we solve these simultaneous equations, and plug our values of l or k into the original equations, only two give integer values for n. ( as required by the question) - the other three are extraneous solutions.
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This was really a nice question. @Soham Karwa , Can you tell me some resources for learning Diophantine equations?
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@Jayakumar Krishnan – Hi Jayakumar!
I would to love to be of service!
When I started to teach myself Number Theory, I too was lost, and didn't really understand how to solve problems.
What really helped me was to dive into a lot of problems, and really understand the solution, even if you get the answer right. That's what helped me!
In terms of resources, I believe Titu Andreescu's book '104 Number Theory Problems' is a great read - it provides a set of in-depth notes to begin with, and then some problems to get your hands dirty!
Also check out AoPs - it's an awesome resource for questions and notes.
I hope that was of help!
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@Soham Karwa – Thanks! That of course..was..of..great..help...Yes..that is a great book...I've heard of it and kinda used it too :D. Well, anything to with Diophantines more ? (104NT has only 2 pages about DIophantines and that too only linear..) Maybe, the other book which ANdreescu has written about NT (theory, structure, examples)....should do! :D..Should you know something...please add in ! (Personal suggestions!_)
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@Jayakumar Krishnan – Hi Jayakumar!
No problem at all - you just reminded me about another of Titu's books (I can't believe I forgot about it, given the nature of the question).
Here it is. - I think there might be a free copy somewhere.
Also, I thought I'd share a few tips about solving diophantine equations:
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@Soham Karwa – Thanks a lot again for replying :D.
Thank you for the explanation.
Hey! @Soham Karwa Thanks a lot for your suggestion of books.
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Let 7 5 n − 5 3 9 = l 2 and 1 2 n − 1 1 9 = k 2 . where n ∈ N
Multiply 7 5 n − 5 3 9 = l 2 by 4 to give 3 0 0 n − 2 1 5 6 = 4 l 2 and 1 2 n − 1 1 9 = k 2 by 2 5 to give 3 0 0 n − 2 9 7 5 = 2 5 k 2 .
Subtract the two new expressions to give 4 l 2 − 2 5 k 2 = 8 1 9 which can be factorised (using the difference of two squares) to give ( 2 l − 5 k ) ( 2 l + 5 k ) = 8 1 9 .
The prime factorisation of 819 is 3 2 ∗ 7 ∗ 1 3 There are five cases to consider. Dealing with the cases (noting that 2 l − 5 k < 2 l + 5 k ) yields that n can only be 2 0 o r 1 2
hence N = 2 0 + 1 2 = 3 2