My problem.
I am trying to fill in the details in @lhf 's outline of an elementary proof (mentioned in Dickson's History of the theory of numbers) that there are no four Squares in arithmetic progression. I have been able to verify all but one part of the outline: I have been unable to check the two equations in red (below). I would appreciate some help with how to prove them.
Proof outline from lhf's answer.
Theorem. Suppose that $\alpha$, $\beta$, $\gamma$ and $\delta$ are rational numbers such that: $$\beta^2-\alpha^2=\gamma^2-\beta^2=\delta^2-\gamma^2. \quad (1) $$ Then $\pm\alpha=\pm\beta=\pm\gamma=\pm\delta$.
I. To prove theorem 1 assume that we are given rational numbers $\alpha$, $\beta$, $\gamma$ and $\delta$ satisfying the hypothesis of the theorem. We may assume that they are relatively prime, non-negative integers.
I understand that this assumption, since
The theorem then states that $\alpha=\beta=\gamma=\delta$. Assume that this is not true. Then we see that $\alpha^2<\beta^2<\gamma^2<\delta^2$.
Show that $\gcd(\alpha,\beta)=\gcd(\beta,\gamma)=\gcd(\gamma,\delta)=1$ and that $\alpha$, $\beta$, $\gamma$, $\delta$ are all odd. Conclude that $\gcd(\beta+\alpha,\beta-\alpha)=\gcd(\gamma+\beta,\gamma-\beta)=\gcd(\delta+\gamma,\delta-\gamma)=2$.
II. Defining positive integers $a,b,c,d$ by: $$2a:=\gcd(\beta+\alpha,\gamma+\beta),\\ 2b:=\gcd(\beta+\alpha,\gamma-\beta),\\ 2c:=\gcd(\beta-\alpha,\gamma+\beta),\\ 2d:=\gcd(\beta-\alpha,\gamma-\beta),$$ show that we have: $$\beta+\alpha=2ab,\\ \beta-\alpha=2cd,\\ \gamma+\beta=2ac,\\ \gamma-\beta=2bd,\\ \color{red}{\delta+\gamma=2bc},\\ \color{red}{\delta-\gamma=2ad}.$$
III. Then $(a+d)b=(a-d)c$ and $(c+d)a=(c-d)b$, and we have the number $\gcd(a+d,a-d)$ and $\gcd(c+d,c-d)$ are both either $1$ or $2$. We can then conclude that $$a+d=mc,a-d=mb,c+d=nb,c-d=na,$$ with $m,n\in\{1,2\}$.
IV. The conclusion in the last step is incompatible with the positivity of $a,b,c,d$.
My attempt.
I am able to verify every step of this argument except for the two equations in red in step II. To prove the first equation in this step, I argue as follows.
Claim. $\beta + \alpha = 2ab$.
Proof. Recall that $\gcd(x, \text{lcm}(y,z)) = \text{lcm}(\gcd(x,y), \gcd(x,z))$. Applying this with $x = \beta + \alpha$, $y = \gamma + \beta$, and $z = \gamma - \beta$, we find that $$\gcd(\beta + \alpha, \frac{1}{2}(\gamma^2 - \beta^2)) = 2ab \quad (2).$$ We claim that $\beta + \alpha$ divides $\frac{1}{2}(\gamma^2 - \beta^2)$. Using (1), observe that $(\beta + \alpha)(\beta - \alpha) = \gamma^2 - \beta^2$. Since $\beta - \alpha$ is even, it is divisible by $2$, so $(\beta + \alpha)\frac{\beta - \alpha}{2} = \frac{\gamma^2 - \beta^2}{2}$.
Thus, the left-hand side of (2) is $\beta + \alpha$, and we get the desired identity.
Using similar arguments, I am able to verify the next three equations in step II. However, the last two equations (in red) don't seem to follow from this method since the numbers $a, b, c, d$ do not involve $\delta$. Multiplying the two red equations gives $2abcd = \delta^2 - \gamma^2$. This also follows from multiplying the first two equations on this line, so proving either of the two red equations should imply the other.
