To find a short way to prove - if $p$ is a prime and $p|a^2+b^2, \space p|b^2+c^2$, then $p|a, p|b, p|a^2-c^2$

Question

To prove, if $p$ is a prime and $p|a^2+b^2, \space p|b^2+c^2$, then $p|a, p|b, p|a^2-c^2$.

$$p|a^2+b^2\\p|b^2+c^2\\ \therefore p|a^2+b^2+b^2+c^2\space\space \text {i.e.}\space \space p|a^2+2b^2+c^2 \space \text {or}\\ p|a^2+b^2-b^2-c^2,\space\space \text {i.e.}\space \space p|a^2-c^2\\ \therefore p|(a+c)\space\space \text {or}\space\space p|(a-c)\\ \Rightarrow p|a+c+a-c \space\space \text {i.e.}\space \space p|2a \space\space \text {i.e.}\space \space p|2 \space\space \text{or} \space \space p|a \\ \text{Similarly}\space\space p|a+c-a+c, \therefore p|2 \space \space \text{or}\space\space p|c\\ \space\\\text{Since},\space \space p|a,\space\\\text{then},p|a^2 \space \space \space\\\text{and},\space \space p|a^2+b^2\\ \space \\\text{Hence}\space\space p|b^2, ( \because p \space\space\text {is prime}\space),\\ \therefore p|b $$..

Does it correctly complete the proof? Or if possible kindly provide a alternative short way to prove. Any suggestion is precious and appreciated.

Answer 1

It is quite easy to see your theorem isn't true. Take $a=c$, and it reduces to

If $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$.

This obviously isn't true, take for example $p=5$, $a=2$ and $b=1$ (and those translate to $(a,b,c,p)=(2,1,2,5)$).