presidentabrahamlincoln 0 Posted February 27, 2017 Report Share Posted February 27, 2017 (edited) Let w, x, y, and z belong to the complete set of integers. If each of wxy + z^2, wxz + y^2, wyz + x^2, xyz + w^2 is divisible by 4, show that w^3 + x^3 + y^3 + z^3 is divisible by 4. Edited February 27, 2017 by presidentabrahamlincoln headline Quote Link to post Share on other sites

0 araver 10 Posted March 4, 2017 Report Share Posted March 4, 2017 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. Quote Link to post Share on other sites

0 CaptainEd 26 Posted March 4, 2017 Report Share Posted March 4, 2017 (edited) 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 March 4, 2017 by CaptainEd Quote Link to post Share on other sites

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## presidentabrahamlincoln 0

Let w, x, y, and z belong to the complete set of integers.

If each of wxy + z^2, wxz + y^2, wyz + x^2, xyz + w^2 is divisible by 4, show that

w^3 + x^3 + y^3 + z^3 is divisible by 4.

Edited by presidentabrahamlincolnheadline

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