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presidentabrahamlincoln

Show 4 divides this algebraic expression.

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I'm sure there is a beautiful algebraic expression out there but ...

Spoiler

This can go the hard way too.

I. Assume x is even, that is x=0 mod 4 or x=2 mod 4.
Either way, x^2 is divisible by 4.
Then wyz=0 mod 4 since wyz+x^2=0 mod 4.
Assume y is odd yields wxz + y^2 odd (since x is even), which contradicts the fact it's divisible by 4.
Similarly all w,y,z are even.
Squares of evens are divisible by 4, hence w^3 + x^3 + y^3 + z^3 is the sum of four numbers all divisible by 4.
q.e.d.

II. Assume x is odd, that is x=1 mod 4 or x=3 mod 4.
Then x^2=1 mod 4 so wyz=3 mod 4.
Clearly none of y,z,w can be even since that contradicts the above.
Let a, b, c be the remainders modulo 4 for y, z, w. Then abc=3 mod 4 and there are only 2 ways that equation holds:
II.A) Two of a,b,c are 1 and the other is 3 since 1*1*3=3 mod 4.
II.B) All of a,b,c are 3 since 3*3*3=3 mod 4.
The other two cases are not possible since 1*1*1=1*3*3=1 mod 4.
Let d be the remainder modulo 4 for x.
Back to the original assumption d=1 or d=3.
II.A) Exactly one of a,b,c is 3 the others are 1.
If d = 1 then because of the symmetry one of the four original products is equal to 1*1*1 + 3^2 = 2 mod 4 contradiction.
If d = 3 then because of the symmetry one of the four original products is equal to 1*3*3 + 1^2 = 2 mod 4 contradiction.
II.B) a=b=c=3
If d = 1 then because of the symmetry one of the four original products is equal to 1*3*3 + 3^2 = 2 mod 4 contradiction.
If d = 3 then w^3 + x^3 + y^3 + z^3=3^3 + 3^3 + 3^3 + 3^3= 1 + 1 + 1 + 1 = 0 mod 4.
q.e.d.

 

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I hope this is clear:

Spoiler

z(wxy+Z^2) = wxyz + z^3 is div 4 
y(wxz+Y^2) = wxyz + y^3 is div 4 
x(wyz+X^2) = wxyz + x^3 is div 4 
w(xyz+W^2) = wxyz + w^3 is div 4 

Now add these four quantities.
sum = 4*wxyz + sum(cubes), and is div 4. Since 4*wxyz is div 4, then so is the sum of the cubes.

Oops! Plasmid beat me to it on the 8:13 version of this puzzle.

Edited by CaptainEd

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