- #1
Albert1
- 1,221
- 0
$m^2-2mn+14n^2=217$
$m,n\in N$
find solution(s) of $(m,n)$
$m,n\in N$
find solution(s) of $(m,n)$
we haveAlbert said:$m^2-2mn+14n^2=217$
$m,n\in N$
find solution(s) of $(m,n)$
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
Albert said:sorry! have some others
kaliprasad said:Yes I missed one more n=4, n-m = 3 giving m = 7, n= 4
Albert said:still one missing
The equation that needs to be solved is $(m^2-2mn+14n^2=217)$.
The variables in the equation are $m$ and $n$.
The method for solving this type of equation is factoring or using the quadratic formula.
Yes, this equation can have multiple solutions for $(m,n)$.
Solving this equation can help in understanding patterns and relationships between variables, which can be useful in various scientific fields such as physics, chemistry, and engineering.