Proof that a cube plus 7 cannot be a square.
                                              By Cino Hilliard

To establish the truth of this proposition we need a couple of lemmas.

Lemma 1:  
All prime factors of numbers of the form  4y^2+1 are of the form 4n+1.
 If a prime p divides a^2+b^2 (but not b), then  (a/b)^2 = -1 (mod p); so -1
is a quadratic residue and it  follows that (-1)^((p-1)/2) = 1 (mod p) and
hence 4 must divide p-1 which implies p = 4n+1 as was desired.
Special thanks to Chris Caldwell for providing this proof.
http://groups.yahoo.com/group/primeform/message/7636

Lemma 2: 
At least one prime factor of numbers of the form 4y^2+3 is of the form 4n+3. 

Case 1. 4y^2+3 is prime.
If 4y^2+3 is a prime p, then p = 4n+1 or p=4n+3. Since p=4n+3 establishes the
Lemma, the case of the form 4n+1 is all we need to dispose of. Assume
4y^2+3 = 4n+1. then we have 4y^2-4n = -2 or  2y^2-2n = -1  implies 2 divides 1,
impossible. Therefore prime p =  4y^2+3 = 4n+3.

Case2: 4y^2+3 has at least 2 prime factors p1 and p2. Since 4y^2+3 is odd, the
the possible forms of p1 and p2 are 4n+1 and 4n+3. Then the combinations of
the forms of p1* p2 = 4y^2+3  are  

1.p1*p2 =  (4n+1)(4m+1) = 16mn+8(m+n) + 1 = 4h+1 = 4y^2+3 => 4(y^2-h) = -2
or 2 divides 1.  So p1 and p2 cannot both be of the form 4k+1.

2. p1*p2 = (4n+1)(4m+3) = 16mn+12n+4m+3 = 4h+3 = 4y^2+3 => y^2 = h for
   some h which is possible. Similarly, p1*p2 = (4n+3)(4m+1) = 4h+3 => y^2 = h. 

3.  p1*p2 = (4n+3)(4m+3) = 16mn+12(n+m) + 9 = 4h+1 = 4y^2+3 => 2 divides 1.

Therefore, p1*p2 = (4n+1)(4m+3) or p1*p2 = (4n+3)(4m+1) are the only possible
forms of p1*p2 allowed and at least one prime divisor of 4y^2+3 is of
the form 4n+3 as was desired. 

We are now ready to prove prove n^3 + 7 != k^2 for all integers n,k.
Assume n^3 + 7 = k^2 for some n,k.

If n is even, say 2h, we test  8h^3 + 7 = k^2. This implies 
8 | k^2 - 7 and since n even => k is odd, say 2s+1. So we have
8h^3  = (2s+1)^2-7 = 4s^2+4s - 6. Dividing by 2, we get
4h^3 = 2s^2 + 2s -3 = form 2z+1 an odd number and a contradiction.
Therefore n is not even.

Then n is odd of the form 2h+1. So (2h+1)^3 + 7 = k^2. If
n is odd, k must be even since (2h+1)^3 = 2z+1+ 7 = 2x+8 = 2H.
Let k =2r. Then (2h+1)^3  + 7 = 4r^2. Adding 1, we have
(2h+1)^3  + 8 = 4r^2+1. Factoring the sum of 2 cubes we have
((2h+1)+2)((2h+1)^2 -2(2h+1)+4) = 4r^2+1
(2h+3)(4h^2+4h+1-4h-2+4) = 4r^2+1
(2z+1)(4h^2+3) = 4r^2+1

Then by Lemma 2, the factor 4h^2+3 of 4r^2+1 has at least one prime factor
that is of the form 4m+3. This contradicts Lemma 1 which states all prime factors
of 4r^2+1 are of the form 4m+1. Therefore, the assumption that n^3+7 = k^2 was
false and there is no cube + 7 that is a perfect square as was desired.