Question

Let a,b,c,d be integers in Z. Find y an integer in Z so that the following expression
\[
a^2 c^2 + 4 a b c^2 + b^2 c^2 + 4 a^2 c d + 16 a b c d + 4 b^2 c d +
a^2 d^2 + 4 a b d^2 + b^2 d^2 + 3 y^2
\]
is a perfect square.

Answer 1

what does Z stand for? it is a subspace??

Answer 2

Z is the set of all integers, positive, negative and zero

Answer 3

well since there are 4 variables present, its most likely a squared 4-nomial?

Answer 4

close ... may have to use the variables as factors ..

Answer 5

a 4-nomial perfect square does expand to 10 terms

Answer 6

so far I have
\[(ac+bc+ad+bd)^2\]
\[a^2 c^2+2 a^2 c d+a^2 d^2+2 a b c^2+4 a b c d+2 a b d^2+b^2 c^2+2 b^2 c d+b^2 d^2\]

Answer 7

\[a^2 c^2+2 a b c^2+b^2 c^2+2 a^2 c d+4 a b c d+2 b^2 c d+a^2 d^2+2 a b d^2+b^2 d^2\]\[a^2 c^2 + 4 a b c^2 + b^2 c^2 + 4 a^2 c d + 16 a b c d + 4 b^2 c d +
a^2 d^2 + 4 a b d^2 + b^2 d^2 + 3 y^2\]

Answer 8

\[a^2 c^2 + 4 a b c^2 + b^2 c^2 + 4 a^2 c d + 16 a b c d + 4 b^2 c d +
a^2 d^2 + 4 a b d^2 + b^2 d^2 + 3 y^2\]

\[(a^2 c^2+2 a b c d + b^2 d^2)\\
~~~~~~~~~~+ 4(a b c^2 + a^2 c d + 3a b c d + b^2 c d + a b d^2) + 3 y^2 \\
~~~~~~~~~~~~~~~~~~~~~~ + (a^2 d^2+2 a b c d + b^2 c^2)\]
let n=(ac+bd), and m = (ad+bc)

\[n^2 + 2(2a b c^2 + 2a^2 c d + 6a b c d + 2b^2 c d + 2a b d^2) + 3 y^2 + m^2\]
what does 2nm equal?

\[2nm=2a^2 c d+2a b c^2+2a b d^2+2b^2 c d\]
that has to equate to \[ 4a b c^2 + 4a^2 c d + 12a b c d + 4b^2 c d + 4a b d^2 + 3 y^2\]


\[~~~ 4a b c^2 + 4a^2 c d + 12a b c d + 4b^2 c d + 4a b d^2 + 3 y^2\]\[-2a b c^2-2a^2 c d-0abcd-2b^2 c d-2a b d^2\]-----------------------------------------\[~~~2a b c^2 + 2a^2 c d + 12a b c d + 2b^2 c d + 2a b d^2 + 3 y^2=0\]

Answer 9

\[
x = a^2 c^2+4 a^2 c d+a^2 d^2+4 a b c^2+16 a b c d+4 a b d^2+b^2 c^2+4 b^2 c d+b^2 d^2\\

y = b c + a d + 4 b d\\

z = a c + 2 b c + 2 a d + 7 b d\\


\]
You can verify that
\[
x+ 3 y^2 =z^2
\]

Answer 10

@guest.100 @Saeeddiscover