Please do try the questions by yourself before resorting to these hints!

1. False, false, and true. Now work out counter-examples for the first two and a proof for the third.
2. It is not good enough simply to show that the given values satisfy the equation. You need to show that every solution is of that form.
   Use Euclid's algorithm to show that 57.17 - 44.22 = 1 and 57.44 - 44.57 = 0.
3. No.
   x = 437k - 308, y = 234 - 332k.
4. x = -31 + 9r - 72s, y = 31 - 8r + 72s, z = -3 - 7s.
   Of course, it is possible to have a different particular solution offset by the complementary function, but there must be two degrees of freedom.
5. 25.
6. a|a and (a,b)|b so a.(a,b)|a.b and a|m. Write a = c.(a,b) and b = d.(a,b), and remember that (c,d) = 1. a|n so write n = e.a = e.c.(a,b), and write n = f.d.(a,b) similarly. Now m = a.b/(a,b) = c.d.(a,b) and n/(a,b) = e.c = f.d. c|n/(a,b) and d|n/(a,b), but (c,d) = 1 so c.d | n/(a,b) and m = c.d.(a,b) | n.
7. Suppose there are only finitely many primes of the form 4k+3. Let p_n be the largest of them and calculate N as in the question. N may be composite but its prime factors are all greater than p_n and so they are all congruent to 1 modulo 4. So their product, N, is also congruent to 1 modulo 4. But it isn’t.
8. The squares modulo 5 are 0, 1 and 4, so there is no α with N(α) = 2.
   4 = 2.2 and 4 = (1 + √5).(-1 + √5).
   Suppose α|2 and α|(1 + √5). Then N(α) = 1 or 4, since it can't be 2. So either α is a unit or 2 and (1 + √5) are both unit multiples of α.
9. Square and add.
   WLoG, suppose (d,e)=k>1. Then k|f as well. d and e can not both be even (else f would be even giving a common factor of 2). d and e can not both be odd (else d²+e² ≡ 2 (mod 4) which could not be f²). d odd and e even makes f odd, so f+d and f-d are both even. 4g² = e² = f²-d² = (f+d)(f-d) = 2h.2i = 4hi. (h,i)|(h+i)=f and (h,i)|(h-i)=d but (f,d)=1. Consider prime factorization of g².
10. Induction on k.
    f_ln = f_n f_(l-1)n+1 + f_n-1f_(l-1)n and use induction on l.
    (f_n,f_n-1) = (f_n-1,f_n-2) = ... =(f₂,f₁) = (1,1) = 1.
    f_m = f_nf_m-n+1 + f_n-1f_m-n. Consider factors and replicate derivation of Euclid’s algorithm.
    f_m | f_mn and f_n | f_mn by above, but (f_m,f_n) = f_(m,n) = f₁ = 1, so f_mf_n | f_mn.
11. Worth thinking about efficiency: only test for odd factors (beyond 2), stop at √n, keep a list of primes and only test for divisibility by them.
12. Define a function that takes two triples (r, s, t) from the extended Euclid’s algorithm and returns the next one.