JOMO 6, Short 10

Write $x^2 - y^2 = 211$ as $(x+y)(x-y)=211$

Since 211 is a prime number, and x,y are positive integers we can say:

$x+y=211$ and $x-y=1$

That's all, our answer is $x+y=211$

Note that the another solution is impossible because x, y could be positive

Notice that the LHS can be factorized into $(x+y)(x-y)=211$ Note that $211$ is prime, hence the only divisors are only $1$ and $211$ . However, the divisors can be both negative or both positive. This gives $2$ possibilities.

Case $1$ : $x+y=211, x-y=1$

Solving this system of equation gives $(x, y)=(106, 105)$ .

Case $2$ : $x+y=-1, x-y=-211$

Solving this system of equation gives $(x, y)=(-106, 105)$ . However, this violates the condition that $x, y\in\mathbb{Z^{+}}$ , hence there is no solution for this case.

Our desired answer is $105+106=\boxed{211}$ .

211 is a prime number. So, it's only divisible by 1 and itself. Now, (x + y)(x - y) = 211 or, x + y = 211/(x - y) Since x and y both are positive integer the sum also can't be fraction and the chance of x - y being 211 is also emitted. So x - y is 1. Then, x + y = 211

As 211 is a prime no. and it has only two factors - 1,211

so now, x^2 - y^2 = (x+y)(x-y)=1*211

So, take (x+y)= 1 and (x-y)= 211

solve and you wil get the values of x and y