We claim $\boxed{Yes}$ Let the $4$-tuple $(a,b,c,d)$ be (O's bday, O's bmonth, M's bday, M's bmonth) So obviously $(11, 6, 22, 12)$ works If there's another tuple, the problem is disproven Then writing algebraically, we have some notes $3d + 3m = 17n$, so $17|m+d$ So now, it becomes trivial, we can take cases for $m+d$ CASE 1: $m+d =17$ So now we can have some subcases, and what we hinge on is that $d = 2m +1$ or $d = 2m-1$ So 3m+1 = 17 or 3m-1 = 17 For the first equation, $m \notin \mathbb{Z}$, so disclude For 2nd case, that's the tuple we already found Now we look for CASE 2: $m+d = 34$ So rewrite it again so $3m-1 = 34$ or $3m+1$ = 34 Only the 2nd equation works, which yeilds $m=11$ But then the month of M's birthday is >12, so this wont work so we have exhausted all our cases, so proven: only tuple is: $(11, 6, 22, 12)$