Are there any integer solutions for $a$, $b$, $c$, $d$ such that $ad^2+{b^2}c=0$

Question

Are there any integer solutions $a$, $b$, $c$, $d$ such that $ad^2+{b^2}c=0$, where $b\neq 0$ and $d\neq 0$? If so, can you please list some examples of them?

Answer 1

Yes there are integer solutions, for example all $(a,b,c,d)\in\mathbb{Z}^4$ with $a=-c, d=\pm b$.

Answer 2

$$ad^2+{b^2}c=0 \Rightarrow \frac{a}{c}=-\left(\frac{b}{d}\right)^2$$ if we take

$a=\pm kb^2$ and $c= \mp kd^2$ we have solutions.

Answer 3

well, obviously $a = -c$ and $b^2 = d^2$.

But less trivially (maybe) for any $a = a'*n^2; n\ne 1$ and $c = -a'$ and $|b| =|dn|$ will do. ($ad^2 + b^2c = a'n^2d^2 + (dn)^2(-a')$.)

So... I dunno $a = 75; b=35; c=-3;d= 7$... $75*7^2 + 35^2*(-3) = 0$.

More thouroughly:

Let $a = \prod p_{a,i}^{n_{a,i}}$ and $d = \prod p_{d,i}^{n_{d,i}}$ be the prime factors of $a$ and $b$ then we $-cb^2=ad^2 =\prod p_{a,i}^{n_{a,i}}\prod p_{d,i}^{2n_{d,i}}$. We can divy up the prime factors between $c$ and $d$ however we like (provided the even/odd parity adds up).

Example: If $a =35=5*7$ and $d = 80=2^4*5$ then $ad^2 =-cb^2 = 2^8*5^3*7$ so I can choose $c= -2^2*5^3*7=3500; d = 2^3=8$ . So $35*80^2 - 3500*8 = 0$. Or $c=-2^8*5*7; d=5$ so $35*80^2 -8960*25 = 0$ (Note: $c$ had to be an odd multiple of 5 and a multiple of 7 and could only have even or 0 powers of 2.)