Can (ab)² + (ac)² + (bc)² be a perfect square when a, b and c are integers?

Question

I am very interested in finding theconditions – if such conditions ever exist – that (ab)² + (ac)² + (bc)² is a perfect square.

Please help. Thanks!

Sonny Torreli, if you put a = 2, b = 4 and c = 5 then you get (ab)² + (ac)² + (bc)² = 564 which is NOT a perfect square! We're dealing here with integers, right? Kindly read the question and details.

Thanks a lot Semyaza for a very comprehensive proof!

By the way a, b and c should be whole integers (non-zero).

I still welcome alternative answers. Thanks!

Thank you very much Duke for contributing a very good approach to my math problem!

Answer 1

Let ω = a² + b² + c² where c = a + b.

(ab)² + (ac)² + (bc)²

= a²(b² + c²) + (bc)²
= a²(b² + c² + a² – a²) + (bc)²
= a²(ω – a²) + (bc)²
= a²ω – a²a² + (bc)²
= a²ω + (bc – a²)(bc + a²)

Consider

a²

= (c – b)²
= c² – 2cb + b²

2a²

= c² – 2cb + b² + a²
= ω – 2cb

Thus ω = 2a² + 2cb or ½ ω = a² + cb.

Recall (ab)² + (ac)² + (bc)² = a²ω + (bc – a²)(bc + a²).

This reduces to

(ab)² + (ac)² + (bc)²

= a²ω + (bc – a²) ½ ω since ½ ω = a² + cb.
= ½ ω [2a² + (bc – a²)]
= ½ ω [a² + bc]
= ½ ω (½ ω) since ½ ω = a² + cb.

THUS (ab)² + (ac)² + (bc)² is a perfect square when c = a + b since (ab)² + (ac)² + (bc)² = ½ ω (½ ω) and
½ ω (½ ω) is a perfect square.

Hope this helps!

Answer 2

Great question, Johnny, congratulations! I haven't encountered such interesting thing for a long time! I'll suggest a solution, not claiming for completeness, the problem being tough.

Now the obvious cases
1 + 4 + 4 = (1*2)^2 + (2*1)^2 + (2*2)^2 = 9 /a=b=1, c=2/
4 + 9 +36 = (1*2)^2 + (1*3)^2 + (2*3)^2 = 49 /a=1,b=2,c=3/
of course satisfy a + b = c as proven in one of the answers, but take
6^2 + 18^2 + 27^2 = 33^2, producing
(2*3)^2 + (2*9)^2 + (3*9)^2 = 1089, here a=2, b=3, c=9 ≠ a+b and there exist infinitely many such solutions. To find them you'll need first to solve the Diophantine equation
x^2 + y^2 + z^2 = R^2 in positive integers, i.e. to find the integer-coordinate points on a sphere /R-radius/. This is a known problem, putting away somewhat lengthy details the solution is given by formulas:
x = |p^2 + q^2 - r^2|
y = 2pr
z = 2qr, here p,q,r are arbitrary integers, 1 or 3 of them odd,
/e.g. p=1, q=3, r=5, so x=15, y =10, z=30, R=35/, or
x = |p^2 + q^2 - r^2|/2
y = pr
z = qr, /p,q,r are arbitrary integers, 0 or 2 of them odd, e.g. p=1, q=2, r=3, so x=2, y=3, z=6, R=7/.

Now, having expressions, producing all triplets (x, y, z), where x^2 + y^2 + z^2 is a perfect square you easily obtain
ab=x, bc=y, ca=z, then
a = sqrt(xz/y), b=sqrt(xy/z), c=sqrt(yz/x), so you must take care xz/y, xy/z and yz/x to be (1) integers; (2) perfect squares. The first is easy - take the least common multiple M of x,y,z then instead of x,y,z take x'=Mx, y'=My, z'=Mz /the sum of their squares is also a perfect square!/, then x'z'/y', x'y'/z' and y'z'/x' will be integers. Go on with the same approach - try to expand with a factor N /suitably chosen - here is the whole thing!/ so that Nx'z'/y', Nx'y'/z' and Ny'z'/x' to be perfect squares - you have now a triplet (a, b, c) as required!

That way you can obtain infinitely many solutions of your nice problem, take this for dessert, obtained as described above:
a=17, b=30, c=85=5*17:
(17*30)^2 + (30*85)^2 +(17*85)^2 = 2975^2

P.S. I tried to suggest an approach how to find numeric solutions, but to find explicit expressions for ALL of them, or to find conditions as you suggest - well, seems difficult!

Answer 3

NO SOLUTION!!!!!

the mddle term should be satisfied
2(ab)(bc) = (ac)^2

2 acb^2=a^2c^2

b = sqrt (ac/2)

(ab)^2 + (ac)^2 + (bc)^2 should be a perfect square

eliminate b in the original equation
a^2 (ac/2) + a^2c^2 + (ac/2) c^2

ac [( a^2 / 2) + ac + c^2/2]

to perfect square
one condition should be

ac = ( a^2 / 2) + ac + c^2/2
0 = a^2 / 2 + c^2/2

which can only be satisfied by a=0 and c = 0

another condition is

a = c [( a^2 / 2) + ac + c^2/2)]
2a = ca^2 + ac^2 + c^3

which does not have any solution

2a < ca^2 ac^2 + c^3
for any positive integer a and c

test for a=1 and c=1 which is the min positive integer
2< 1+1+1

Answer 4

middle term = 2 (ac)(bc) = (ac)^ 2
. . . . . 2 ab c^2 = a^2 c^2
. . . . . .2 b = a . . . . . substitute the given equation
(2b b )^2 + (2b c)^2 + (bc) ^2
4 b^4 + 4 (bc)^2 + (bc) ^2
4 b^4 + 5 (bc)^2
. . . . hard

Answer 5

surely
wat kiddish stuff u asking man?
put a,b,c anything from R

u get a perfect square
even if u put a=b=c=iota (sqrt(-1))
even then
('coz iota squared= -1)