Question

1) a and b are natural numbers such that 2013 + a² = b² then minimum value of ab is             
2) Let x,y,z be three positive integers such that x + y + z = 10 .The max value for xyz + xy + xz + yz is                 

Answer 1

1)
       b² - a² = 2013
       (b-a) (b+a) = 2013 = 3 * 11 * 61      prime factors.
 
b-a is smaller than b+a as and b are natural numbers.  
Also b>a as RHS is positive.
 
Possibilities:  b - a = 3   or   11 or   33 
     correspondingly  b+a  is  671  or  183  or  61

      then  b = 337 and a = 334   or   b = 97 & a = 86     or   b = 47 & a = 14

Product ab is least when a = 14 and b = 47  and is 658
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2)
           S= x + y + z = 10      x, y, z ∈ N - { 0 }   positive integers

A quick way is: 
     The given expression to be maximized :  V = xyz + xy + yz + zx

     S and V are symmetric in x, y and z.  So maximum V is obtained when x = y = z = c.
     Then c = 10/3 = non-integer.  The possible values of x,y, and z are around 3.3.

Some possible combinations are: 3, 3, 4 ;;; 2, 4, 4  ;;;  5, 3, 2 ;; 

x,y,z are : 3, 3, 4   =>  V = 69     this is the maximum.
x, y , z =  2, 4, 4    =>  V = 64 
x, y, z =  5, 3, 2    =>   V = 61
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  ANother way:     1<=  x, y, z <= 8

  V = 9 x y + 10 x + 10 y - x² y  - x y²  - x² - y²,       
 you can find  dV/dx and dV/dy  - partial derivatives and equate to 0.  Then you find x = y =  3.3.  at this value V is maximum.

So choose x, y, and z nearest to this value 3.3 and you get the maximum.

You can alternatives like above.    x,y,z are  3,3,4 or 5,3,2 or 4,2,4 .   For the combination 3,3,4 the value of the expression V is maximum and is 69.