Quadratic Cubes

Number Theory Level 4

n 2 + n + 1 = m 3 \Large n^2+n+1=m^3

If n n and m m are integers, how many solutions are there to the equation above?


The answer is 4.

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Garrett Clarke by Garrett Clarke , 25, USA

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1 solution

n ( n + 1 ) = ( m − 1 ) ( m 2 + m + 1 ) = ( m − 1 ) { ( m − 1 ) ( m + 2 ) + 3 ) } g c d ( m − 1 , ( m − 1 ) ( m + 2 ) + 3 ) = 1 , 3 ∵ g c d ( a , a m + b ) = g c d ( a , b ) ∴ n = 3 ( m − 1 ) a n d n + 1 = m 2 + m + 1 3 ∴ m 2 − 8 m + 7 = 0 S o l v i n g t h e q u a d r a t i c m = 1 , 7. F o r m = 1 , n = 0 , − 1. F o r m = 7 , n = − 19 , 18. 4 s o l u t i o n s . n(n+1)=(m-1)(m^2+m+1)=(m-1)\{(m-1)(m+2) +3)\}\\ gcd(m-1,(m-1)(m+2)+3)=1,3~~~~\because gcd(a,am+b)=gcd(a,b)\\ \therefore~n=3(m-1)~~and~~n+1=\dfrac{m^2+m+1} 3\\ \therefore ~m^2-8m+7=0\\ Solving~the~quadratic~m=1, 7.\\ For~m=1,~n=0,~-1.~~~For~m=7,~n=-19,~18.\\ 4~solutions. \\ The method above I learn from a similar problem "Find Prime Numbers!" by Satyajit Mohanty. There was a solution by ? ? ? ? ? b u t i t i s n o t t h e r e n o w ! ! \text{The method above I learn from a similar problem } \\ \text{"Find Prime Numbers!" by Satyajit Mohanty. There was a solution by} \\ ?????~but ~it~ is~ not ~there~ now!!

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????? was me :)) my solution was wrong.

Kazem Sepehrinia - 5 years, 10 months ago

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Thanks you. Can you please give details about why your solution was wrong? I have mentioned the rule from web on which the solution depends. gcd(a.b)=gcd(a, am+b).

Niranjan Khanderia - 5 years, 10 months ago

While you did get 4 solutions, those aren't the 4 solutions. The four points, presented in the form (n,m), should be (0,1), (-1,1), (18,7), and (-19,7). You're close though!

Garrett Clarke - 5 years, 10 months ago

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I might have not pressed SUBMIT or there might be some problem with Brilliant.org. I had replied to you and made correction, but it is not there now!Any way. My replay: Thank you and you are correct. While copying from my notes, I made the mistake which I am correcting.

Niranjan Khanderia - 5 years, 10 months ago

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No problem, thanks for the solution!

Garrett Clarke - 5 years, 10 months ago

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