I solved this sum by trial and error method.

no of cabbage is directly proprtinal to the area.ni of cabage is C and side of square patch is X. Therefore

C kX^2

C has to be a perfect square.

now the area is increased by sum amount say Y and no of cabage is increased by 211

C+211 K(X+Y)^2

C+211 also has to b a perfect square

that means our job is to find the two perfect square whose difference is 211.

Consider the square root of first no as a+d and square root of second no is a+e where a is the base part same for the both the nos and only e is greater than d.

now as the diff between their square is 211 therefore

(a+e)^2 - (a+d)^2 211

a^2+2ae+e^2-a^2-2ad-d^ 2 211

2a(e-d)+(e^2-d^2) 211

now i have used trail and error to split 211 as 200+11 where 11 is e^2-d^2 and 200 is 2a(e-d) and i got e 6 d 5 a 100 therefore apples in his patch are 106*106 and apples in her patch is 105*105

Here'a the clearest solution i could think of :

With the simple matrix funda the cabbages would have been arranged in a square matrix sort of fashion ( obviously because we have a square patch ). So let us assume that the matrix this year has 'n' rows and 'n' columns. Obviously the cabbages last year would form a matrix with say 'x' rows and 'x' columns where x<n. So the simplest equation formed is : (n^2)-(x^2) 211. Assuming that the value of n is 20 ( by a simple assumption u can also start with a value such as 16 or 17 the only point is that the 'square of that number' - 211 should give u a perfect square ) we get 400 ( i.e. 20^2 ) - 211 289 which is a perfect square for 17. Hence we get the answer as 400 cabbages this year after having 289 cabbages last year.

P.S : the answer i.e. 106*106 cabbages this year after 105*105 cabbages last year is also correct ( and can be achieved with the same simple equation ) but has an assumption that the two perfect squares should be consecutive which is not stated in the problem.