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I started like this :

$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$

but it's leads to nowhere. can you help please ?

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marked as duplicate by GNUSupporter 8964民主女神 地下教會, Delta-u, Namaste, Dave, user99914 May 2 '18 at 15:40

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Suppose that this Diophantine equation has an integral solution $(a,b,c)\neq(0,0,0)$. Then, $a\neq 0$ and $b\neq 0$, whence $a^2-1\geq 0$ and $b^2-1\geq 0$. If $a$ or $b$ is even, then $a^2-1\equiv 3\pmod{4}$ or $b^2-1\equiv3\pmod{4}$. Hence, $\left(a^2-1\right)\left(b^2-1\right)=c^2+1$ is divisible by a prime natural number $p\equiv 3\pmod{4}$. Thus, $c^2\equiv -1\pmod{p}$, which is a contradiction as $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=-1$. Therefore, $a$ and $b$ must be odd. Thus, $$c^2=a^2b^2-a^2-b^2\equiv1-1-1=-1\pmod{8}\,,$$ which is a contradiction.

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  • $\begingroup$ thanks Batominovski for you answer, just a short question how did you get : $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=-1$ $\endgroup$ – xAminex May 16 '16 at 13:35
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This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)

assuming the equation is

$a^2+b^2+c^2=t^2$ ,∀ t=ab

all integer positive solution given by

$a=(l^2+m^2−n^2)/n , b=2l, c=2m$ ,and $t=(l^2+m^2+n^2)/n$

Then

$2l(l^2+m^2−n^2)=l^2+m^2+n^2$

one can find that $(l^2+m^2−n^2)(2l−1)=0$

suppose $2l−1=0$ refused

then

$l^2+m^2−n^2=0⇒a=0 $ ,and $ab=2n$

but $ab=0⇒n=0⇒l^2+m^2=0⇒b=0,c=0$ therefore the only solution is (0,0,0)

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