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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.

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    $\begingroup$ I have checked APs with first term and difference up to $1000$ with no examples found, which (amazingly to me) suggests there are no solutions. If this is indeed the case, I expect this to be an enormously difficult problem - already nonexistence of four squares in an AP is nontrivial, and this question is asking about integer points on a certain arithmetic surface. AFAIK, the theory of points on surfaces is very poorly understood, much less than for curves. $\endgroup$ – Wojowu Apr 22 at 11:37
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    $\begingroup$ WLOG the four terms be $a\pm3d, a\pm d$ The product is $$(a^2-9d^2)(a^2-d^2)=a^4-10a^2d^2+9d^4=(a^2-5d^2)^2-(4d^2)^2$$ This needs to be perfect square $$(a^2-5d^2)^2-(4d^2)^2=r^2$$ $\endgroup$ – lab bhattacharjee Apr 22 at 11:51
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    $\begingroup$ @Peter I think you're thinking of the lack of nontrivial solutions to $x^2+y^4=z^\color{blue}{4}$. In fact, since fourth powers aren't of the form $4k+2$, they're always the difference of two squares. $\endgroup$ – J.G. Apr 22 at 17:21
  • $\begingroup$ @Peter Two examples of what J.G. said in comment: $312^2+5^4=313^2$ and for case of even fourth power, $30^2+4^4=34^2.$ $\endgroup$ – coffeemath Apr 22 at 17:40
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    $\begingroup$ Yes, I remembered it wrong. $\endgroup$ – Peter Apr 22 at 18:22
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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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Suppose $a < b < c < d$ are positive integers in arithmetic progression and $abcd=r^2$. If $a,b,c,d$ have a common factor, and their highest common factor is $m$, then $a/m, b/m, c/m, d/m$ will also be positive integers in arithmetic progression and their product $(a/m)(b/m)(c/m)(d/m)$ will also be a square: $(r/m^2)^2$. So to prove impossibility it suffices to prove it for the case where it is assumed that $gcd(a,b,c,d) = 1$.

Since there cannot be four squares in arithmetic progression, at least one of the four integers must have a prime factor, say $p$, to an odd power. In order that the product of the integers be a square, at least one other of the four integers must also have the factor $p$ to an odd power. For the special case in which these two integers are adjacent in the arithmetic progression, we can then reason as follows. Suppose the two integers are $a$ and $b$ (the argument can readily be adapted to the other adjacent pairs), with $a=Ap$ and $b=Bp$. Then $b-a=p(B-A)$ and therefore:

$$c-b = p(B-A)$$

$$c = b + p(B-A) = Bp + p(B-A) = p(2B-A)$$

and also:

$$d-c = p(B-A)$$

$$d = c + p(B-A) = p(2B-A) + p(B-A) = p(3B - 2A)$$

Thus $p$ is a common factor of $a,b,c,d$ contradicting our assumption.

Perhaps someone can extend this reasoning to address cases where the two integers with the factor $p$ to an odd power are not adjacent.

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    $\begingroup$ If $a=pA$ and $c=pC$ have $p$ to an odd power then the step size in the progression is $p\frac{C-A}{2}$ and $b$ and $d$ are also multiples of $p$, unless $p=2$. Similarly if $a$ and $d$ are multiples of $p$ unless $p=3$. $\endgroup$ – Solomonoff's Secret Apr 22 at 21:04
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    $\begingroup$ @Solomonoff'sSecret Yes, those are the hard cases. $\endgroup$ – Adam Bailey Apr 22 at 21:15
  • $\begingroup$ Following up, your answer would be crisper if it ended by reducing the original problem to that of whether $y^2 = x(x+2)(x+4)(x+6)$ or $x(x+3)(x+6)(x+9)$. $\endgroup$ – Solomonoff's Secret Apr 23 at 12:35

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