Let x, y be integers. what possible values can x^2 + y^2 take in z4

0, 1 or 2.

Step-by-step explanation:

An integer x in Z4 is either equal to [0], [1], [2] or [3] (as Z4 is made only of the remainders we can get when dividing an integer by 4).

If x was equal to [0] in Z4, then x^2 = [0]*[0]=[0] in Z4.

If x was equal to [1] in Z4, then x^2 = [1]*[1]=[1] in Z4.

If x was equal to [2] in Z4, then x^2 = [2]*[2]=[4]=[0] in Z4 (as 4 and 0 are the same in Z4, given that both numbers leave a remainder of 0 when divided by 4).

If x was equal to [3] in Z4, then x^2 = [3]*[3]=[9]=[1] in Z4 (as 9 and 1 are the same in Z4, given that both numbers leave a remainder of 1 when divided by 4).

Therefore, in Z4 x^2+y^2 is either a sum of the form [0]+[0], or [0]+[1], or [1]+[0], or [1]+[1], which means we can only get either [0], [1] or [2].

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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