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 integer solutions where \(m, n \in \mathbb{N}\). The discussion identifies that the equation can be rearranged and solved using algebraic techniques. Specific integer pairs \((m, n)\) that satisfy the equation are derived through systematic substitution and testing of values. The final solutions are confirmed as valid within the constraints of natural numbers.

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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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