Solving $(m,n)$ for $m^2-2mn+14n^2=217$

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Discussion Overview

The discussion centers around finding natural number solutions $(m,n)$ for the equation $m^2-2mn+14n^2=217$. Participants are exploring potential approaches to solve this equation.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants restate the equation and the requirement for $m$ and $n$ to be natural numbers.
  • Others suggest that there may be multiple approaches to finding solutions, although specific methods have not yet been detailed.
  • A later reply indicates that further elaboration on the equation is forthcoming, but no specific claims or solutions have been presented yet.

Areas of Agreement / Disagreement

Participants have not yet reached any consensus on specific solutions or methods, and the discussion remains unresolved.

Contextual Notes

There are no explicit assumptions or definitions provided that could limit the scope of the discussion, but the nature of the equation suggests potential complexities in finding solutions.

Albert1
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$m^2-2mn+14n^2=217$
$m,n\in N$
find solution(s) of $(m,n)$
 
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Albert said:
$m^2-2mn+14n^2=217$
$m,n\in N$
find solution(s) of $(m,n)$
we have

$(m-n)^2 + 13n^2 = 217$
so $13n^2 < 127$ or $n^2 < 17 ( 13 * 17 = 221)$
trying n = 1,2,3,4,5 we get
n = 3 and m-n = 10 so m = 13 , n =3 is the solution and no other
 
Last edited:
kaliprasad said:
we have

$(m-n)^2 + 13n^3 = 217$
so $13n^2 < 127$ or $n^2 < 17 ( 13 * 17 = 221)$
trying n = 1,2,3,4,5 we get
n = 3 and m-n = 10 so m = 13 , n =3 is the solution and no other
sorry! have some others
 
Albert said:
sorry! have some others

Yes I missed one more n=4, n-m = 3 giving m = 7, n= 4
 
kaliprasad said:
Yes I missed one more n=4, n-m = 3 giving m = 7, n= 4
still one missing
 
Albert said:
still one missing

I should have taken |m-n| =3 giving m = 1 and n = 4
 

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