In order to factorize the integer $N = 67591$, choose a factor base $\{2,3,5\}$ and four congruences: $24256^2 \equiv 2^9 \cdot 3^4(mod\ N)$; $59791^2 \equiv 2^2 \cdot 3^4\cdot 5^2(mod\ N)$; $23541^2 \equiv 2^3 \cdot 3^2(mod\ N)$; $29396^2 \equiv 2^9 \cdot 3^4(mod\ N)$. Then how to determine the prime factors of $N$?
Could someone give me some hints about this problem? I have no idea at all.
If you have any instance of $a^2\equiv b^2\pmod N$, this means $N$ divides $(a+b)(a-b)$. If you're lucky, it does not divide purely one of the two factors and so $\gcd(a-b,N)$ is possibly a proper divisor of $N$. (If you are not lucky, the gcd is either $1$ or $N$, shit happens).
From the data points, you have, you can infer $$24256^2\equiv 29396^2\pmod N $$ $$ 59791^2\equiv 90^2\pmod N$$ $$ (24256\cdot 23541)^2\equiv (2^6\cdot 3^3)^2\pmod N$$ $$ (29396\cdot 23541)^2\equiv (2^6\cdot 3^3)^2\pmod N$$
and perhaps some of these may give you a non-trivial proper divisor.
The first two relations are "immediate" fro the data, the other are obained by combining suitable of the results in such a way that the exponents of the small primes $2,3,5$ become all even.