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Find the positive integers \(n\) for which \(n+9,16n+9\) and \(27n+9\) are all perfect squares.

Note by Akshat Sharda
6 months, 4 weeks ago

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Let \(n+9=x^2\) \(16n+9=y^2\) and \(27n+9=z^2\) now multiply the first equation with 16 and subtract it with the second. You'll be left with \(16x^2-y^2=135\) Similarily with the first and third mulitiply with 27, you'll be left with \(27x^2-z^2=234\). For the last multiplication step multiply the second with 27 and and third with 16 you'll be left with \(27y^2-16z^2=99\) now \(27y^2-16z^2+16x^2-y^2=234=27x^2-z^2\) By arranging that we get \(26y^2=15z^2+11x^2\) Now the obvious solution is \(x=y=z=1\) which leaves us with a contradiction that no such integers n exist. However I am not sure that there are no more solutions to this equation(I haven't come across that field of study yet!) Good luck I hope I helped Dragan Marković · 6 months, 3 weeks ago

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@Dragan Marković Nice sol.+1..how did u think of that? Rishabh Tiwari · 6 months, 3 weeks ago

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@Rishabh Tiwari Well actually with elimination. I have 2 appoaches to such problems. Either factorization or eliminating the unknowns. Factorization failed horribly i got nothing from it so i tried this way and ended up with the final equation. So i guess the factroization didn't fail that horribly? And this method was way shorter to write ;) Dragan Marković · 6 months, 3 weeks ago

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@Dragan Marković Hey! See this, \(n=280\) satisfies. Akshat Sharda · 6 months, 3 weeks ago

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@Akshat Sharda Wow did some more analysis. If we can prove that \(7y=5z+2x 《》25y=30z-55x\) we would get 280 for n. This is intriguing 7=5+2 and 25=55-30!!!! Dragan Marković · 6 months, 3 weeks ago

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@Akshat Sharda It is the solution to my equation as well!!! I don't know how to get 17 67 and 87 though we haven't done that in school yet. Maybe someone like Calvin Lin could help us with that? Dragan Marković · 6 months, 3 weeks ago

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