This problem was proposed by me.

Let $E$ the symmetric of $A$ in $B$ then $ACE $ is right angled hence $D,C,E$ are collinear and $DAE$ is isoceles thus $d^2-c^2=AC^2=(d-c)^2+4b^2 \implies 2dc=4b^2 $ since $c\wedge b=1$ then $b^2 $ divide $ 2d$ therefore $2d=2^i.b^2$ and the result follows. RH HAS

PROF65 wrote: Let $E$ the symmetric of $A$ in $B$ then $ACE $ is right angled hence $D,C,E$ are collinear and $DAE$ is isoceles thus $d^2-c^2=AC^2=(d-c)^2+4b^2 \implies 2dc=4b^2 $ since $c\wedge b=1$ then $b^2 $ divide $ 2d$ therefore $2d=2^i.b^2$ and the result follows. RH HAS Thus not $d^2-c^2=AC^2=(d-c)^2+4b^2$ but $d^2-c^2=AC^2=-(d-c)^2+4a^2$ And it implies $2d^2=4a^2+2dc.$

So $$d^2=2a^2+dc.$$And call it $(*).$ Let $v_p(n)$ be the power of prime divisor $p$ in positive integer $n.$ Suppose both $d$ and $2d$ are not perfect squares. Then there exist an odd prime divisor $p$ such that $v_p(d)=t$ is an odd integer. Hence, $$v_p(d^2)=2t \geq t+1 \textup{ and } v_p(dc)=v_p(d)+v_p(c)=t+v_p(c) \geq t.$$Thus, with $(*)$ we have $$2v_p(a)=v_p(a^2)=v_p(2a^2) \geq t.$$Because $v_p(a)$ is a non-negative integer and $t$ is an odd integer, $$v_p(2a^2)=2v_p(a) \geq t+1 \textup{ and } p|a.$$So $$t+v_p(c)=v_p(dc) \geq t+1.$$And we get $$v_p(c) \geq 1 \implies p|c.$$Contradiction! Thus either $d$ or $2d$ is a perfect square.

If we prove that d=1?then??

Rizsgtp wrote: If we prove that d=1?then?? Sorry, but I can't figure out what you mean.