<s>[INST] can the product of four positive integers in A.P. be a square? Title says it. It's known no four squares are in arithmetic progression, but it's asking less for their product to be square.
I've tried various things like hunting to make each of two subproducts a square, or one $ua^2$ and the other $ub^2,$ no success for an example. On the other hand I can't prove it's impossible. Thanks for any response.
 [/INST] Solution 1
We can see that the roots of $P(a)=a(a+d)(a+2d)(a+3d)$ are $0,-d,-2d,-3d$. The values $0,-3d$ are roots for the quadratic $x^2+3dx=0$ while the other pair are the roots for the quadratic $x^2+3dx=-2d^2$. Thus, we can see that all four are roots of $(x^2+3dx+d^2)^2-(d^2)^2$. Since this is a biquadratic monic polynomial and so is $P(x)$ and they share the same roots:
$$a(a+d)(a+2d)(a+3d)=(a^2+3ad+d^2)^2-(d^2)^2$$
Let $a(a+d)(a+2d)(a+3d)=k^2$. Then, we have: $$k^2+(d^2)^2=(a^2+3ad+d^2)^2$$
Due to homogeneity, one can set $\gcd(a,d)=1$. One can prove that $d$ cannot be even (Refer solution 2). Thus, we have:
$$d^2=m^2-n^2 \space ; a^2+3ad+d^2=m^2+n^2 \implies a(a+3d)=2n^2$$
$$a(a+d)(a+2d)(a+3d)=2mn \implies (a+d)(a+2d)=\frac{2mn}{2n^2}=\frac{m}{n}$$
However, $\frac{m}{n}$ cannot be an integer unless $n=1$ since $\gcd(m,n)=1$. Then, $a(a+3d)=2$ wouldn't be possible, since $a(a+3d) > 1 \cdot 4 =4>2$. Thus, there are no solutions for the product of the four distinct naturals in AP to be a perfect square.
Solution 2
Assume $\exists$ $(a,b,c,d)$ s.t. the quadruple form an AP whose product is a perfect square. We can see that if $\gcd(a,b,c,d)=g$ then:
$$abcd=m^2 \implies \frac{a}{g} \cdot \frac{b}{g} \cdot \frac{c}{g} \cdot \frac{d}{g} = \bigg(\frac{m}{g^2}\bigg)^2$$
Now, set $(w,x,y,z)=(\frac{a}{g},\frac{b}{g},\frac{c}{g},\frac{d}{g})$ and $n=\frac{m}{g^2}$. Thus, we have another quadruple having the same property, additionally containing terms whose $\gcd$ is $1$.
Now, by Fermat's $4$ square theorem, it is not possible for all of $(w,x,y,z)$ to be a perfect square. Now, consider the term s.t. $\exists$ prime $p$ dividing it and the power of $p$ dividing it is odd. 
If $p$ divides the adjacent term (WLOG let the term and its adjacent term be $w$ and $x$), then:
$$p \mid w \space ; p \mid x \implies p \mid (x-w) \implies p \mid d \implies p \mid y \space ; p \mid z \implies \gcd(w,x,y,z) \geqslant p>1$$
This is a clear contradiction as we know that the $\gcd$ of the quadruple is $1$.
Now we have the two cases:
$(1)$ $p \mid w$ and $p \mid z$
$(2)$ $\bigg(p \mid w$ and $p \mid y\bigg)$ or $\bigg(p \mid x$ and $p \mid z\bigg)$
If $p$ divides the $1$st and $3$rd terms only, or the $2$nd and $4$th terms only (WLOG $w$ and $y$), then:
$$p \mid 2d \space ; p\nmid d \implies p=2$$
Now, this means that $d$ is odd which shows that $x$ and $z$ are odd. Moreover, $z-x = 2d \implies 4\nmid (z-x)$. Thus, one of $x$ and $z$ is $1 \pmod{4}$ and the other is $3 \pmod{4}$. But the term that is $3 \pmod{4}$ cannot be a square as $l^2 \neq 4k+3$. This shows that the term that is $3 \pmod{4}$ shares a prime factor with another term. It cannot be the other one of $x$ and $z$ as it would then make both terms even. It also cannot be an adjacent term. Thus, the terms $w,z$ which are farthest apart have to share a prime factor. This means that $(2) \implies (1)$.
One can similarly show that $(1) \implies (2)$ by showing that one of the other terms is $2 \pmod{3}$ and $l^2 \neq 3k+2$. Thus, we have $(1) \iff (2)$. However, since one of them is true, both of them are true.
This tells us that our quadruple has to be of the form: $$(w,x,y,z)=(6a^2,b^2,2c^2,3d^2)=(q,q+r,q+2r,q+3r)$$
This gives us two equations:
$$a^2+d^2=c^2$$
$$(2a)^2+d^2=b^2$$
We know that $a$ is even and $d$ is odd from our $(w,x,y,z)$ quadruple work. Since $(a,b,c,d)$ are pairwise relatively prime, we have:
$$a^2+d^2=c^2 \implies a=2mn \space ; d=m^2-n^2$$
Now, we have:
$$(2a)^2+d^2=b^2 \implies (m^2-n^2)^2+(4mn)^2=b^2$$
Thus, we have:
$$m^2-n^2 = x^2-y^2 \space ; 4mn=2xy$$
$$m^2+y^2=x^2+n^2 \space ; \frac{2m}{y}=\frac{x}{n}=2t \space ; t \in \mathbb{Q}$$
$$y^2(t^2+1) = n^2(4t^2+1) \implies \frac{4t^2+1}{t^2+1} = z^2 \space ; z \in \mathbb{Q}$$
We know that $t = \frac{m'}{y'}$ (The simplified form of $\frac{m}{y}$). This shows that $(2m')^2+(y')^2$ and $(m')^2+(y')^2$ are squares. We required solutions for $(2X)^2+Y^2$ and $X^2+Y^2$ to be squares. But-
$$(X,Y)=(2mn,m^2-n^2) \implies (X',Y')=(m',y')$$
Here, $(X',Y')$ is a smaller solution. By descent, we will never get the smallest solution since $X$ can always decrease. This is only possible if there are no solutions (Reductio Ad Absurdum).
Hence, there are no $4$-term APs with their product as a square (if all terms are distinct).
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