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# Need solution

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

Note by Akshat Sharda
11 months, 2 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 · 11 months, 2 weeks ago

Nice sol.+1..how did u think of that? · 11 months, 2 weeks ago

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 ;) · 11 months, 2 weeks ago

Hey! See this, $$n=280$$ satisfies. · 11 months, 2 weeks ago

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!!!! · 11 months, 2 weeks ago