Suppose there exists $a,b,c$ that satisfy the condition and $abc\neq 0$. We can assume $a,b,c$ are natural number. Let's get a pair of $(a,b,c)$ which $b$ is the smallest. Let) $d=pa^2-3b^2$ $d^2-8b^4=c^2$ $8b^4=(d+c)(d-c) \cdots $(1) Let) $x=gcd(c,d), e=\frac{c}{x}, f=\frac{d}{x}$ $(gcd(e,f)=1)$ by (1) $x|b $ Let) $g=\frac{b}{x}$ $8x^2g^4=(f+e)(f-e)$ $2|f+e, 2|f-e$ $2x^2g^4=\frac{f+e}{2} \frac{f-e}{2}$ as $gcd(\frac{f+e}{2}, \frac{f-e}{2})=1, \{\frac{f+e}{2}, \frac{f-e}{2}\}=\{2h^2, j^2\}$ for some natural numbers $h,j$ we can know $gcd(h,j)=1, 2\nmid j $ $f=2h^2+j^2, hj=xg^2$ by definition of $d$, $pa^2=d+3b^2=xf+3x^2g^2=x(2h^2+j^2+3hj)$ $pa^2g^2=hj(2h+j)(h+j) \cdots$ (2) as$ j$ is odd number, $gcd(j, 2h+j)=1 $ therefore $h,j,h+j,2h+j$ is pairwise coprime we can know that one of them is a form of $p*(square)$, while other are $square$ now casework I) $h+j$ is $p*(square)$ $(square)+(square)=h+j=h+j=p*(square)$ by modulo 4, we can get $h,j,h+j$ has to be even number. contradiction. II) $2h+j $ is $p*(square)$ $(square)+(square)=(h+j)+h=2h+j=p*(square)$ likewise I), contradiction. III) $j$ is $p*(square)$ as $j$ is odd number, $j\equiv 3 (mod 8)$ $h+3\equiv h+j (mod 8)$, so $h\equiv 1 (mod 8), h+j\equiv 4 (mod 8)$ $2h+j\equiv 5(mod 8)$, so contradiction. IV) $h$ is $p*(square)$ by similar modulo work as III), we can get $h$ is even number where $j,h+j,2h+j$ is odd number. Let) $h=4pk^2, j=\gamma ^2, h+j=m^2, 2h+j=n^2$ as $m,n$ is odd number, $\frac{n-m}{2} \frac{n+m}{2}=\frac{-(h+j)+(2h+j)}{4}=\frac{h}{4}=pk^2 \cdots (2)$ as $gcd(m,n)=1$, $\{\frac{n-m}{2}, \frac{n+m}{2}\}=\{p\alpha^2, \beta^2\}$ for some natural numbers $\alpha, \beta$. therefore $n=|p\alpha^2-\beta^2|, m=p\alpha^2+\beta^2$ by (2) $k=\alpha \beta$ therfore $\gamma^2=j=h+j-h=n^2-4pk^2=(p\alpha^2-\beta^2)^2-4p\alpha^2 \beta^2=p^2\alpha^4-6p\alpha^2\beta^2+\beta^4 \cdots$ (3) $b=xg\ge\sqrt{xg^2}=\sqrt{hj}\ge \sqrt{h}=\sqrt{4p\alpha^2\beta^2}\ge \beta \cdots$ (4) by (3), (4), contradiction Therefore contradiction on all cases, so Claim is right.

Suppose there exists $a,b,c$ that satisfy the condition and $abc\neq 0$. We can assume $a,b,c$ are natural number. Let's get a pair of $(a,b,c)$ which $b$ is the smallest. Let) $d=pa^2-3b^2$ $d^2-8b^4=c^2$ $8b^4=(d+c)(d-c) \cdots $(1) Let) $x=gcd(c,d), e=\frac{c}{x}, f=\frac{d}{x}$ $(gcd(e,f)=1)$ by (1) $x|b $ Let) $g=\frac{b}{x}$ $8x^2g^4=(f+e)(f-e)$ $2|f+e, 2|f-e$ $2x^2g^4=\frac{f+e}{2} \frac{f-e}{2}$ as $gcd(\frac{f+e}{2}, \frac{f-e}{2})=1, \{\frac{f+e}{2}, \frac{f-e}{2}\}=\{2h^2, j^2\}$ for some natural numbers $h,j$ we can know $gcd(h,j)=1, 2\nmid j $ $f=2h^2+j^2, hj=xg^2$ by definition of $d$, $pa^2=d+3b^2=xf+3x^2g^2=x(2h^2+j^2+3hj)$ $pa^2g^2=hj(2h+j)(h+j) \cdots$ (2) as$ j$ is odd number, $gcd(j, 2h+j)=1 $ therefore $h,j,h+j,2h+j$ is pairwise coprime we can know that one of them is a form of $p*(square)$, while other are $square$ now casework I) $h+j$ is $p*(square)$ $(square)+(square)=h+j=h+j=p*(square)$ by modulo 4, we can get $h,j,h+j$ has to be even number. contradiction. II) $2h+j $ is $p*(square)$ $(square)+(square)=(h+j)+h=2h+j=p*(square)$ likewise I), contradiction. III) $j$ is $p*(square)$ as $j$ is odd number, $j\equiv 3 (mod 8)$ $h+3\equiv h+j (mod 8)$, so $h\equiv 1 (mod 8), h+j\equiv 4 (mod 8)$ $2h+j\equiv 5(mod 8)$, so contradiction. IV) $h$ is $p*(square)$ by similar modulo work as III), we can get $h$ is even number where $j,h+j,2h+j$ is odd number. Let) $h=4pk^2, j=\gamma ^2, h+j=m^2, 2h+j=n^2$ as $m,n$ is odd number, $\frac{n-m}{2} \frac{n+m}{2}=\frac{-(h+j)+(2h+j)}{4}=\frac{h}{4}=pk^2 \cdots (2)$ as $gcd(m,n)=1$, $\{\frac{n-m}{2}, \frac{n+m}{2}\}=\{p\alpha^2, \beta^2\}$ for some natural numbers $\alpha, \beta$. therefore $n=|p\alpha^2-\beta^2|, m=p\alpha^2+\beta^2$ by (2) $k=\alpha \beta$ therfore $\gamma^2=j=h+j-h=n^2-4pk^2=(p\alpha^2-\beta^2)^2-4p\alpha^2 \beta^2=p^2\alpha^4-6p\alpha^2\beta^2+\beta^4 \cdots$ (3) $b=xg\ge\sqrt{xg^2}=\sqrt{hj}\ge \sqrt{h}=\sqrt{4p\alpha^2\beta^2}\ge \beta \cdots$ (4) by (3), (4), contradiction Therefore contradiction on all cases, so Claim is right.