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

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The equation \(m^2 - 2mn + 14n^2 = 217\) is analyzed for natural number solutions \(m\) and \(n\). Participants explore various values for \(n\) and calculate corresponding \(m\) values to find valid pairs. The discussion emphasizes the need for systematic testing of natural numbers to identify all possible solutions. Some users suggest using algebraic manipulation to simplify the equation further. Ultimately, the goal is to determine all pairs \((m, n)\) that satisfy the equation.
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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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