Question

Let a,b,c,d,e be integers in \( \bf Z\). Find m an integer in \( \bf Z\) so that the following expression
\[
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+3 m

\]
is a perfect square.

Answer 1

expression. I did not type the p

Answer 2

Is there a method for solving this? Or just choosing numbers for a,b,c,d,e so that its a perfect square?

Answer 3

Actually, I still do not know how to do it.

Answer 4

I was able to find numerically values of m for any given values of a,b,c,d .

Answer 5

Not sure. But a^2c^2, a^2d^2, b^2c^2, b^2d^2 seems to suggest (ac + ad + bc + bd)^2 and that may be a starting point?

Answer 6

That is what I thought, but could not go further.

Answer 7

for a=1,b=2,c=3,d=4, all these values of m less than 1000 will do the job
{{4, 961}, {25, 1024}, {69, 1156}, {92, 1225}, {140, 1369}, {165,
1444}, {217, 1600}, {244, 1681}, {300, 1849}, {329, 1936}, {389,
2116}, {420, 2209}, {484, 2401}, {517, 2500}, {585, 2704}, {620,
2809}, {692, 3025}, {729, 3136}, {805, 3364}, {844, 3481}, {924,
3721}, {965, 3844}}
the second terms gives you the perfect square.
I was able to find m for any particular choices of a,b,c,d

Answer 8

I do point out that
\[a^2 c^2+4 a^2 c d+a^2 d^2+4 a b c^2+16 a b c d+4 a b d^2+b^2
c^2+4 b^2 c d+b^2 d^2= \\
\left(a^2+4 a b+b^2\right) \left(c^2+4
c d+d^2\right)\]

Answer 9

I now found a solution
\[
m=2 \left(-c d \left(a^2+4 a b+b^2\right)-a b \left(c^2+4
c d+d^2\right)+6 a b c d\right)
\]
would work. I will explain how.

Answer 10

\[a^2 c^2+4 a^2 c d+a^2 d^2+4 a b c^2+16 a b c d+4 a b
d^2+b^2 c^2+4 b^2 c d+b^2 d^2+3 m= \\6 \left(-c d
\left(a^2+4 a b+b^2\right)-a b \left(c^2+4 c
d+d^2\right)+6 a b c d\right)+\\a^2 c^2+4 a^2 c d+a^2
d^2+4 a b c^2+16 a b c d+4 a b d^2+b^2 c^2+4 b^2 c
d+b^2 d^2=\\a^2 c^2-2 a^2 c d+a^2 d^2-2 a b c^2+4 a b c
d-2 a b d^2+b^2 c^2-2 b^2 c d+b^2 d^2=\\(a-b)^2 (c-d)^2\]

Answer 11

@tHe_FiZiCx99
@ranga
@satellite73

Answer 12

That equation for m will identify just one of the many values of m for a given a,b,c,d and most likely a negative integer for m for a=1,b=2,c=3,d=4.
Similarly, my earlier suggestion of (ac + ad + bc + bd)^2 which simplifies to (a+b)^2(c+d)^2 which when expanded and equated to the original expression will identify another value for m.
But is there any way to come up with a general formula that will identify ALL or at least a few different values of m that will make the expression a perfect square for a given a,b,c,d?

Answer 13

Ahhh o: ranga is right though. One of many.

Answer 14

Good point @ranga

Answer 15

@ranga, Are you sure that the m you get from your computation is an integer?
What I got from solving for m the equation
\[
a^2 c^2+4 a^2 c d+a^2 d^2+4 a b c^2+16 a b c
d+4 a b d^2+b^2 c^2+4 b^2 c d+b^2 d^2+3
m=\\(a+b)^2 (c+d)^2
\]
gives m
\[
m=-\frac{2}{3} \left(a^2 c d+a b c^2+6 a b c
d+a b d^2+b^2 c d\right)
\]
Which is not an integer in general

Answer 16

@eliassaab, I didn't go through the computation to find m and I didn't notice until now that m has to be an integer too.

Answer 17

Yes, m has to be an integer and this is where the difficulty is, otherwise if you call w
\[
w=a^2 c^2 + 4 a b c^2 + b^2 c^2 + 4 a^2 c d + 16 a b c d + 4 b^2 c d +
a^2 d^2 + 4 a b d^2 + b^2 d^2\\
w+ 3 m = x^2\\
m = \frac {x^2-w}{3}
\]

Answer 18

@ranga

Answer 19

True.
For a given set of integers a,b,c,d "w" can be computed and w will be an integer.
So the general solution for m, with a little bit of trial and error, could be:
m = (N^2 - w) / 3
where N is any integer AND (N^2-w) is a multiple of 3.