$\frac{2002}{a^2+c^2+3}$ is a perfect square. We only need to check the 8k+3, 4, 7, 0 factors of 2002. They are, 7, 11, 91, 143. Then it remains simple checking.

siuhochung wrote: $\frac{2002}{a^2+c^2+3}$ is a perfect square. We only need to check the 8k+3, 4, 7, 0 factors of 2002. They are, 7, 11, 91, 143. Then it remains simple checking. How come $\frac{2002}{a^2+c^2+3}=b^2-1$ would be a perfect square ?

Oh well, we indeed get \[2002= (b^2 - 1)(a^2 + c^2 + 3)\] One of the factors on the right hand side must be even because the left hand side is, but it can't be the first one since then $b$ must be odd and the right hand side will be a multiple of 8 while the left hand side obviously isn't. So the second factor must be even and the first one must be odd. But it is impossible for the second factor to be equal to 2 (mod 4) because the square residues (mod 4) are 0 and 1. So there are no solutions.

Yes, you're right! Indeed, the official solution is esentially the same: factorize the expression, and then compare the form of the factors in order to obtain a contradiction. The worst thing about this pretty equation is that it succumbs to a straightforward modulo 4 approach... You even don't need to factorize it

shyong wrote: siuhochung wrote: $\frac{2002}{a^2+c^2+3}$ is a perfect square. We only need to check the 8k+3, 4, 7, 0 factors of 2002. They are, 7, 11, 91, 143. Then it remains simple checking. How come $\frac{2002}{a^2+c^2+3}=b^2-1$ would be a perfect square ? oops sorry, I got a mistake in calculations.

We can rewrite the equation as $(a^{2}+c^{2}+3)(b^{2}-1)=2002$ Now, the divisors of $2002$ are $ 1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002 $, and the only one that can be written as $b^{2}-1$ with $b$ integer is $143$ what leaves us with $a^{2}+c^{2}=11$ and this last equality has no integer solutions for both $a,c$.

Note: the post just above mine is probably the best-written solution to this problem. Just in case you needed help with this problem, like I did.

Note: the post just two above mine is probably the best-written solution to this problem. Just in case you needed help with this problem, like I did.

It can be done checking mod 4, you get that a^2b^2+b^2c^2-b^2-c^2-a^2==1 mod 4, then b^2 can't be 1 mod 4 and neither 0 mod 4, so there aren't any solutions