Find values $a$ and $b$ which satisfies $a^2 \equiv b^2\ (mod\ N)$ by factoring number $N = 52907$. Use given identities:
$$399^2 \equiv 480\ (mod\ N)\ \ \ 480 = 2^5 * 3 * 5$$ $$763^2 \equiv 192\ (mod\ N)\ \ \ 192= 2^6 * 3$$ $$773^2 \equiv 15552\ (mod\ N)\ \ \ 15552= 2^6 * 3^5$$ $$976^2 \equiv 250\ (mod\ N)\ \ \ 250= 2 * 5^3$$
What I did is calculate:
$$ 399^2 * 773^2 * 976^2 \equiv 480*1552*250\ (mod\ 52907) $$ $$ \equiv (2^5 * 3 * 5)*(2^6 * 3^5)*(2 * 5^3)\ (mod\ 52907) $$ $$ = 2^{12}* 3^6 * 5^4$$ $$ = (2^{6}* 3^3 * 5^2)^2 $$ $$ \equiv 43200^2\ (mod\ 52907) $$
Next:
$$399 * 773 * 976 \equiv 36829\ (mod\ 52907)$$
And: $gcd(52907, 43200-36829) = 277$. Thus $52907 = 277 * 191$.
But I am not sure if this is correct at all? Any help how to solve this kind of problem would be appreciate...
$52907 = 191*277$
$a^2 \equiv b^2 \mod 52907$ means $a^2 - b^2 = k*52907$
So $(a + b)(a-b) = k*191*277$
Several possible solutions. If we ignore the trivial $a = n; b = \pm n$ for any integers $n$ (maybe by assuming $a$ and $b$ are positive and $a \ne b$) we can still find several where $(a+b)$ and $(a-b)$ are factors of $k*191*277*$.
The most apparent is $a+b = 277$ and $a - b = 191$ so $2b = 86; b=43;a=234$. But there are infinitely many more.
(Example $a+b = 52907; a - b = anything;$ so $a= 907; b =52000$. Or $a+b = 3*191; a-b=277;$ so $a$ and $b$ equal... whatever.)
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I'm honestly not sure what this exercise wants you to do or what the point is but I do notice that.
$192 = 2^6*3$ so $192*9^2 = 2^6*3^5 = 15552$
So $(763*9)^2 \equiv 192*9^2 = 15552 \equiv 773^2 \mod 52907$.
So $a = 763*9; b= 773$ then $a^2 \equiv b^2 \mod 52907$ answers your exercise. However I must confess I see absolutely no point to this exercise and I have no idea what anyone can learn from it.