Proof that every odd integer is a difference of two squares How can I use a direct proof to show that every odd integer is the difference of two squares?
 A: I certainly agree that Dan Shved's answer is entirely correct.  Moreover, the OP's idea that "guessing is not allowed" in mathematics is a bit confused; after a certain point in mathematics, guessing is usually not allowed unless you can support your guessing with correct reasoning; barring the latter doesn't make much sense, since a description of the thought processes by which one arrived at a proof has no part in the formal proof process and in practice is rarely included in the "informal" proofs we write for others to read.
On the other hand, I think we would learn much better from each other if we did include more information about the thought processes by which we arrived at our arguments.  In a lot of cases the student is apparently supposed to perceive the intuition or idea behind the argument just by reading an unmotivated proof.  However, in my experience this is very difficult for many students, who often (i) lack the mathematical sophistication to perform this process on their own; (ii) lack the depth of understanding of the specific subject matter being presented necessary to analyze an argument for the ideas behind it; and/or (iii) simply are not aware that they could/should be doing this.  Certainly this skill is one that I developed only in graduate school and beyond.
Let me provide a more motivated answer to this question.  This was something that I came up with only when teaching courses a few years past my PhD: I really don't think it is something that students perceive in the standard "Well, take $a = k+1$ and $b = k$" answer.
We are trying to solve that for every odd integer $a$, the equation $x^2-y^2 = a$ has a solution in integers $x$ and $y$.  When I see $x^2-y^2$, the first thing I think of is that, unlike say $x^2+y^2$, it factors as $(x+y)(x-y)$.  So we are trying to solve $(x+y)(x-y) = a$.  Well, golly, maybe we can just take $a = x+y$ and $1 = x-y$.  (Yes, this is also a guess!  Indeed most mathematical thought involves some guesswork in this sense.  But to me it feels like a very small, and natural, guess, compared to just guessing the entire solution.)  Then we get the system 
$x+y = a$
$x-y = 1$
to solve.  Well, okay: adding the equations gives $2x = a+1$, so $x = \frac{a+1}{2}$, and thus $y = \frac{a-1}{2}$.  This makes sense for any $a$ if $x$ and $y$ are allowed to be rational numbers.  But if we want them to be integers, we need $a+1$ and $a-1$ to be even, i.e., $a$ to be odd.  Notice that if we write $a = 2k+1$ we have recovered the "guess" in Dan Shved's answer.
But let's not stop just by answering the original question: rather, let's play around to see what else we can get.  More generally, if $a = bc$ we can try taking
$x+y = b$
$x-y = c$  
and then $x = \frac{b+c}{2}$, $y = \frac{b-c}{2}$ is a solution as long as $b$ and $c$ have the same parity.  (This is now a more verbose, motivated form of lab bhattacharjee's answer.)  Which integers $a$ can we write as $bc$ with $b$ and $c$ of the same parity?  We already did the case in which $a$ is odd.  If $a$ is even, then $b$ and $c$ both need to be even, so $a$ must be divisible by $4$.  Thus if $a$ is $2$ modulo $4$ then the equation $x^2-y^2 = a$ cannot have an integer solution.  However, if $a$ is divisible by $4$ we can take $b = \frac{a}{2}$ and $c = 2$.  Thus we've proved:

An integer $a$ is of the form $x^2-y^2$ for integers $x,y$ if and only if $a$ is odd or divisible by $4$.  

There are further lessons to be drawn from this problem, e.g.
1) A binary quadratic form $ax^2+bxy+c^2$ sometimes factors as a product of two linear forms, and if so this factorization is helpful in determining which integers it represents.
(We could go on to understand when these factorizations take place in terms of the polynomial $ax^2+bx+c$ having rational roots and thus a square discriminant.)  
2) Sometimes it is useful to make a change of variables.  E.g. asking which integers are of the form $q(x,y) = xy$ would be a silly question: all of them, obviously.  But taking $(x,y) \mapsto (x+y,x-y)$ brings us from this silly form to our form.  The change of variables is invertible if dividing by $2$ is not an issue: when it is, there is some further subtlety, but we can work it out.  
So showing what's behind your "Just observe that..." statements can be very educational.  
A: Just observe that $2k+1 = (k+1)^2 - k^2$.
UPDATE: On the subject of guessing. I feel something needs to be said about it. I also feel that it would be even better if someone else pitched in, maybe someone who has more experience in logic or teaching or teaching logic.
My hint expands to a proof that goes more or less like this:

We need to prove that for any integer $k$ integers $a$ and $b$ exist such that $a^2 - b^2 = 2k+1$. Take $a = k+1$ and $b=k$ and observe that indeed $2k+1 = a^2 - b^2$. QED.

The "guessing" happens when we say out of nowhere "Take $a = k+1$ and $b=k$". My point is, and I cannot stress this enough, that this kind of guessing is absolutely OK when you prove the existence of something. The best way to prove existence of an object with desired properties is to construct such an object, and guessing is the simplest one-step construction.
The foundations for this can be found in the axioms of predicate calculus. One of the axiom schemas of predicate calculus says:
$$
( \forall x \mathcal{B}(x) ) \to \mathcal{B}(t),
$$
where $x$ is any variable, $\mathcal{B}$ is any wff (well-formed formula) and $t$ is any term that is substitutable for $x$ in $\mathcal{B}$ (roughly, no variables mentioned in $t$ must be bound by quantifiers when $t$ is substituted for free occurrences of $x$ in $\mathcal{B}$).
From this axiom schema and some basic propositional calculus, one can get:
$$
\mathcal{B}(t) \to (\exists x \mathcal{B}(x)).
$$
It follows that if you prove $\mathcal{B}(t)$, you also prove $\exists x \mathcal{B}(x)$.
This can be applied directly to our problem. Look at the wff $\mathcal{B}(k, a, b)$: "$2k+1 = a^2 - b^2$". We see that statement $\mathcal{B}(k, k+1, k)$ says "$2k+1 = (k+1)^2-k^2$", and this is considered known (can be proved). Then, applying the rule above twice, we see that statement $\exists a \exists b \mathcal{B}(k, a, b)$ can also be proved, and that is exactly what we need.
Now, the exact wffs can vary depending on your framework (you could be working in set theory, or in some flavor of arithmetic, for instance), but the message stays the same. If you have proved that some specific object is "good", then you have also proved that at least one "good" object exists.
A: HINT:
$$a\cdot b=\left(\frac{a+b}2\right)^2-\left(\frac{a-b}2\right)^2$$
So, we need $a\pm b$ to be even $\implies a,b$ must be of same parity
For odd number $2c+1=(2c+1)\cdot1$
