Quadrilaterals with integer sides and square integers.?

I think the answer is that there are none. However, I do not have a proof, so at this point all I can report is that if such a quadrilateral exists then at least one of its sides has length greater than 10000, because my program looked at the others. Also, if the condition on perimeter is replaced by it being 2 times a square, then there are numerous examples. I will think some more about this.

@gianlino: so far I don't see how to parameterize 1+a^2 = b^2 + c^2. Do you know a way? As for the factor of 2 situation, one can find a family of quadrilaterals with sides [1, (m^2-1)^2-m^2+2, 2*m^2-3, (m^2-1)^2-m^2] and square integer area m^2*(m^2 - 2)^2. These have two opposing right angles and perimeter 2 times a square, namely 2 (m^2-1)^2. If you stretch by a factor of 2 the area will multiply by 4 and the perimeter by 2 so both will be squares, giving infinitely many of the desired type of quadrilateral, but with minimum edge 2. They will all be dissimilar, since the gcd of the sides is square free. But I don't see how to use this to get an obstruction to quads with minimum side 1.

@gianlino: very clever. I will see what I can do with that.

I've turned the problem into a system of Diophantine equations, which might help us spot an "algebraic obstruction". So far I've only used it to show that the semi-perimeter is even, hence has an odd number of factors of 2. Perhaps with more work this can give a contradiction. Rita the dog's replacement condition where twice the perimeter, i.e. 4 times the semi-perimeter, is a square would force an even number of factors of 2 in the semi-perimeter, so at least there's some hope with this line of thought.

Let a, b, c, and d be side lengths (where I'll take a = AB's length, etc.). Let p = a+b+c+d be the perimeter, s=p/2 be the semi-perimeter, and K be the area. Note that the quadrilateral is cyclic (since opposite angles are supplementary; see my first reference under "Characterizations"), so from Brahmagupta's formula,

K = sqrt((s-a)(s-b)(s-c)(s-d))

From the Pythagorean Theorem, we find

a^2 + b^2 = c^2 + d^2

which moreover forces the angles at B and D to be right (again see my first reference, this time under "Angle Formulas"). By assumption a=1 and the other sides are distinct integers. In all,

b, c, d are distinct integers greater than 1

p = 1+b+c+d is a perfect square

s = p/2

K = sqrt((s-1)(s-b)(s-c)(s-d)) is a perfect square

...[or K = (b + cd)/2 is a perfect square]

1 + b^2 = c^2 + d^2

As noted, we may replace Brahmagupta's formula with K = (b + cd)/2 by just considering triangles [though it also falls out of the algebra after some effort, as it must]. From the above assumptions, you can construct a cyclic quadrilateral with the original properties, so they characterize the geometric problem completely. Yet again, we've reduced an interesting problem to solving Diophantine equations.

Take the final equation mod 2 to get

1 + b = c + d (mod 2)

p = 1 + b + c + d = 0 (mod 2)

Since p is a perfect square, it is divisible by 2 an even number of times, hence by 4, so p/2 = s is an even integer and is divisible by 2 an odd number of times.

If AB = a, BC = b, CD = c and DA = 1, and let AC = x, then by Pythagoras,

1 + c^2 = x^2

Then, using parameters of Pythagorean triples, either :

m^2 - n^2 = 1, 2mn = c and m^2 + n^2 = x

But, m^2 - n^2 = 1 is impossible in integers m, n. In fact, 5 is the smallest it can be.

or :

m^2 - n^2 = c, 2mn = 1 and m^2 + n^2 = x

But, 2mn = 1 is impossible.

Thus, the answer is none.

EDIT: Oh! dear, you're so right gianlino. How did I think the answer would be so easy?