Do four natural numbers exist which satisfy these constraints?

Question

Do four natural numbers $a,b,c$ and $d$ exist such that the following three conditions are true?

$$a^2+b^2+2d^2=c^2$$ $$\sqrt{a^2+d^2}\in\mathbb{N}$$ $$\sqrt{b^2+d^2}\in \mathbb{N}$$

Answer 1

The smallest example is $a=9, b=16, c = 25, d = 12$. Of course, any integer multiple of this will do. There are other solutions as well.

Answer 2

You want $(a^2+d^2)+(b^2+d^2)=c^2$ and also $a^2+d^2=x^2$ and $b^2+d^2=y^2$.

The solutions of $x^2+y^2=c^2$ are $x=2mnk,y=(m^2-n^2)k,c=(m^2+n^2)k$ where $(m,n)=1$ and $m>n$.

Now we have the equations $a^2+d^2=(2mnk)^2$ and $b^2+d^2=((m^2-n^2)k)^2$.

Do the same and see what happens:)

Answer 3

For any Pythagorean triple $(x,y,z)$ (such that $x^2+y^2=z^2$) denote $$ a=x^2,\\ b=y^2,\\ c=z^2,\\ d=xy. $$

Then $$ a^2+b^2+2d^2 = a^2+b^2+2ab = (a+b)^2=z^4=c^2, $$ $$ a^2+d^2=x^4+x^2y^2=x^2(x^2+y^2)=(xz)^2, $$ $$ b^2+d^2=y^4+x^2y^2=y^2(y^2+x^2)=(yz)^2. $$