can the product of four positive integers in A.P. be a square?

Solution 1:

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).

Solution 2:

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.