An enthusiastic engineer is working on a new Substrate service and thought it is a good idea to implement his version of Schnorr in python3. If you look at the implementation, you may quickly find this might not be the best idea. He event shared to of his signatures!
Update: Yes, the signatures would need to be different, so now it is fixed. Due to special circumstances and chaos, end your solution directly to six in this case.
Flag: you need to find the private key and send it to the flag API, eg CCTF{0o012341234123412341234}
Public Key: 22341107127610550988374035416706441983513459134252228703346605310354889960934500320108205356826404540644249667697298119618250632592931925656991079615926519282687033429512433610343417192289951593171044318706324610168336794983141253851498113254802167480992453266779242096141421575367353312940295740783752468397
Signature 1: (36010016668617097361277333486252221367265386372767366241079427549116533976925351949437714336011792447352848903525769796203416477947326301260607372451782302092572340963064832955737062891765900287304143970127836793719469048561866053685078001905256175778382947915983235836297415450822689242179636102066373966373, 50925148014331406608576349676505764411987614062018614609074827785267540101)
Signature 2: (37181834035952841320766994973344251562596307096325322886712428598593638443440020653320423482108409728759660891541934153105832691335163152265963115894979733985718598848614255129823826377778782269776034131770967469426675073752452874570612647653331940339555999589562003027756352079721278998817934727554592540718, 50925148014331406608576349676505764411987614062016283217379060290906714245)
//Type: Cryptography, Schnorr
//Author: six
six
7a95a813da