Can this be proved wrong?

Question

Make a rectangle ABCD where AB & CD are the longer sides (lenghts)of the rectangle ABCD.Connect the diagonal AC.Extend BC upto E & connect ED such that the triangle ABE is a right angled triangle.Connect ED.Here if CD is given in integer values then we will not get AE in integer values if ED is having integer values or if AE is having integer values then we will not get ED in integer values.
Can this be proved wrong?

Here the two shorter sides (breadth) of the rectangle ABCD , AD & BC is also having integer values.

Answer 1

If you don't assume that BC or CE are integers then there are plenty of counterexamples.

OK now all sides of the initial rectangle are integers. That makes it more challenging!

Ok Here is a counterexample. I am not saying there are no easier ones.: set p= 13x15x17. Then

p^2 = (13x13x15)x(15x17x17)
=(13x13x17)x(15x15x17).

From Pythagoras theorem you get 2 right triangles with sides
(p,a,b) and (p,a',b') where b-a=(13x13x15), b+a=15x17x17,
b'-a'=13x15x15 and b'+a'=13x17x17. So a=15x(289-169)/2 = 900.
and
a' = 13x(289-225)/2 = 416. So if you set AB = CD =p =3315 and
AD=BC= 900-416=484, then by extending BC so that BE= 900 and CE= 416, AE and DE will have integer values
b=900+2535=3435 and b'= 416+2925=3341.

What you asked for is true however if the number of factors in the decomposition of AB into primes is at most 2. A counterexample needs at least 3.

Answer 2

it can be proved wrong by finding a counter example.

Let AB = CD = 3.
Let AD = BC = 2.
Let BE = 4.

Then AB = 3.
Then AE = 5.

Both CD and AE are integer.

So your theory is proved wrong.

I see your additional comment, and it doesn't change the proof.

Answer 3

Yes when you give high numbers for the sides