a little easy rsa

RSA, but d = p. d is also small, so you’re seemingly supposed to use some kind of Coppersmith thing (the flag text indicates this as well), but I think the author just messed up.

Full buffer for my solve, including the notes I write in comments as I tend to “think by typing” while working out the algebra or Euler hacks (two dead end thoughts and then the correct one):

n = 73105772487291349396254686006336120330504972930577005514215080357374112681944087577351379895224746578654018931799727417401425288595445982938270373091627341969888521509691373957711162258987876168607165960620322264335724067238920761042033944418867358083783317156429326797580005138985469248465425537931352359757
e = 4537482391838140758438394964043410950504913123892269886065999941390882950665896428937682918187777255481111874006714423664290939580653075257588603498124366669194458116324464062487897262881136123858890202346251370203490050314565294751740805575602781718282190046613532413038947173662685728922451632009556797931
c = 14558936777299241791239306943800914301296723857812043136710252309211457210786844069103093229876701608756952780774067174377636161903673229776614350695222134040119114881027349864098519027057618922872932074441000483969146246381640236171500856974180238934543370727793393492372475990330143750179123498797867932379

# e*p = k*phi + 1
# e*p = k*(n-p-q+1) + 1
# k*(q-1) - 1 = 0 (mod p)
# k == 1/(q-1) (mod p)
#
# e*p - 1 = k*phi
# e/n = k/p - k/n - k/p^2 + (k+1)/np
# e/n = (k*p-k)/p^2 - k/n + (k+1)/np

# e*p - 1 == k*phi
# e*n == q (mod phi)
# n == p+q-1 (mod phi)
# n-e*n == p-1  (mod phi)

print(bytes.fromhex(hex(pow(c,n-e*n+1,n))[2:]))

Solved in minutes.