# Need solution

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

Note by Akshat Sharda
2 years, 3 months 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

- 2 years, 3 months ago

Hey! See this, $$n=280$$ satisfies.

- 2 years, 3 months 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!!!!

- 2 years, 3 months ago

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?

- 2 years, 3 months ago

Nice sol.+1..how did u think of that?

- 2 years, 3 months 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 ;)

- 2 years, 3 months ago