The positive integers a, b and c satisfy a^-2+b^-2=c^-2. WHat is the sum of all possible value of "a" such that "a" is less than or equal than 100
\[\frac{ 1 }{ a ^{2} }+\frac{ 1 }{ b ^{2} }=\frac{ 1 }{ c ^{2} }\]
But your question makes no sense. What is the sum of all possible value of "a" such that "a" is greater or equal than 100? The sum of "a" is just "a". If that's it, then great... a must be greater than or equal to 100... a >=0
1/a^2 = 1/c^2 - 1/b^2 =(b^2 - c^2) / b^2 c^2 a= bc/sqrt(b^2 - c^2) now,, is RHS an integer ever ?
I guess yes if b^2-c^2=x^2
but it should also divide bc..
Yep...... And also it must be greater or equal to 100
yep..
we maybe just need a proof that no such no. exists..
Well I guess there are infinte number of solution........ One of them is: a=100 , b=75 , c=60
Next one will be a=120 , b=5*18 , c=4*18
I guess u mean a is less than or equal to 100
|dw:1350114527248:dw| where x is a set of odd numbers greater than or equal to 3 and y can be any integer such that a is positive and a>=100.......... AND THIS GIVES INFINITE SOLUTION OF a.
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