Be cool. Don't be fooled

• The largest $x + y$ sum of any number will always be when x is the number itself and y is 1 , as the all other sums are substantially lower.
• According to this, we have the largest sum is $(x = 3721) + (y = 1) = 3722$ .

(x+y)^2 = (x-y)^2 + 4xy.
For (x+y) to be max, (x-y) should be max because 4xy is fixed. For (x-y) to be max, x needs to be max and y needs to be min. Minimum possible y is 1 => x=3721. Ans. = 3721+1=3722.

3721 = (61)(61) = (1)(3721) = (3721)(1)

Then

x = 61, y = 61

or

x = 1 , y = 3721

or

x = 3721 , y = 1

Then

The maximum value of x + y is 3722

Let x=u+v, y=u-v, xy>0, thus x and y are both positive or both negative, but since we are finding the maximum value of x+y, we can assume that x and y are both positive, hence u>0 and u>v.

And we can assume x>y without loss of generality aince x and y can be switched freely. So we can assume u>v>0.

x+y=2u is directly proportional to u, so the larger the u, the larger the x+y

3721=xy=(u-v)(u+v)=u^2-v^2

And now we want to make u as large as possible, but u^2-v^2 is a constant, larger the v^2, the larger the u^2.=> larger the v, the larger the u.

So we need to make v as large as possible.

v=(x-y)/2

So the larger the value of x- y, the larger the value of x+y. xy=3721, x=3721 , y=1 give us the largest value of x-y, so largest value of x+y=3721+1=3722

Only have two solutions for $x + y$ : $122$ or $3722$

Then the answer is $\boxed {3722}$

The lowest value we can consider of x or y is 1. The x or y=3721. Then x+y=3721+1=3722

The answer had to be a positive odd number... First thing tried was root. 61 is a prime. Factors of 3721 are 1, 61, 3721.

$ab=cd\quad \wedge \quad \sqrt { { \left( a-b \right) }^{ 2 } } <\sqrt { { \left( c-d \right) }^{ 2 } } \quad \Rightarrow \quad a+b<c+d\\ ab=cd\quad \wedge \quad d=1\quad \Rightarrow \quad c=ab\\ ab=cd\quad \wedge \quad d=1\quad \Rightarrow \quad \sqrt { { \left( c-d \right) }^{ 2 } } =\sqrt { { \left( ab-1 \right) }^{ 2 } } \\ ab=cd\quad \wedge \quad d=1\quad \Rightarrow \quad \sqrt { { \left( a-b \right) }^{ 2 } } \le \sqrt { { \left( c-d \right) }^{ 2 } } \\ ab=cd\quad \wedge \quad d=1\quad \wedge \quad \sqrt { { \left( a-b \right) }^{ 2 } } <\sqrt { { \left( c-d \right) }^{ 2 } } \quad \Rightarrow \quad a+b<ab+1\\ 3721=xy\\ xy+1=3722\\$

let $f(x)=x+y$

$f(x)=x+\frac {3721} {x}$

$f'(x)=1-\frac {2(3721)} {x^2}$

When $f'(x)=0$ , $x= 86.27,-86.27$

For $f'(x)>0$ at $x>86.27$ , then $f(x)$ continues to increase. But what is the max value of $x$ such that $x>86.27$ and is a factor of $3721$ ? $3721$ .

Thus, $f(x)=3722$ .

Also, $f(x)$ continues to increase as $x$ moves away from $86.27$ towards $0$ . What is the smallest factor of $3721$ then? $1$ .

Thus, $f(x)=3722$ .