First assume that

We will prove by contradiction that there are no $ a,b,c,d$ that are four distinct positive integers in arithmetic progression such that $ abcd$ is a perfect square. Suppose $ a = x + k_1y, b = x + k_2y, c = x + k_3y, d = x + k_4y$ such that $ k_1\not = k_2\not = k_3\not = k_4$ $ abcd = (x + k_1y)(x + k_2y)(x + k_3y)(x + k_4y)$ $ = x^4 + (\sum_{i = 1}^{4} k_i)x^3y + (\sum_{i\not = j} k_ik_j)x^2y^2 + (\sum_{i\not = j \not = m} k_ik_jk_m)xy^3 + k_1k_2k_3k_4y^4$ From $ abcd$ is perfect square, so let $ x^4 + (\sum_{i = 1}^{4} k_i)x^3y + (\sum_{i\not = j} k_ik_j)x^2y^2 + (\sum_{i\not = j \not = m} k_ik_jk_m)xy^3 + k_1k_2k_3k_4y^4$ $ = (x^2 + axy + by^2)^2$ for some $ a,b\in \mathbb{R}$ From $ (x^2 + axy + by^2)^2 = x^4 + 2ax^3y + (a^2 + 2b)x^2y^2 + 2abxy^3 + b^2y^4$, we get: $ \sum_{i = 1}^{4} k_i = 2a$ $ \sum_{i\not = j} k_ik_j = a^2 + 2b$ $ \sum_{i\not = j\not = m} k_ik_jk_m = 2ab$ $ k_1k_2k_3k_4 = b^2$ From Viete's Formula, we get $ k_1,k_2,k_3,k_4$ are roots of the polynomial: $ P(z)=z^4 - 2az^3 + (a^2 + 2b)z^2 - 2abz + b^2 = 0$ $ (z^2 - az + b)^2 = 0$ which, obviously, can be seen that there must be some $ i\not = j$ such that $ k_i = k_j$ so a contradiction occurs. Therefore, there are no $ a,b,c,d$ that are four distinct positive integers in arithmetic progression such that $ abcd$ is a perfect square.

Assume there exists a nonempty set of solutions. Then there exists a solution with least value of $ abcd$ in this set, and we look at this solution. It is clear that there is no common divisor $ h>1$ such that $ a=ha', b=hb', c=hc', d=hd'$, or else $ (a',b',c',d')$ is another solution, because $ a'b'c'd'=\frac{abcd}{h^4}$ and thus is still a square if $ abcd$ is, and $ a'b'c'd'<abcd$, contradicting the choice of $ (a,b,c,d)$. Now, we write $ b=a+k,c=a+2k,d=a+3k$. The above is equivalent to $ \text{g.c.d.}(a,k)=1$. It follows that if $ p>3$ is prime, $ p$ divides at most one of $ a,b,c,d$, because $ p|a+ik, p|a+jk \Rightarrow p|(j-i)k \Rightarrow p|k (\because |j-i|\leq 3) \Rightarrow p\not| a \Rightarrow p\not|a+ik$ contradiction. Thus $ a,b,c,d$ are all squares multiplied by 2s and 3s. If we don't mind our squares containing extra factors of 2 and 3, it follows that they are all of one of the forms $ m^2,2m^2,3m^2,6m^2$. Suppose one of $ a,b,c,d$ is of the form $ 3m^2$ or $ 6m^2$. Then, for $ abcd$ to be a square, and thus divisible by an even power of 3, another of them must be of the form $ 3m^2$ or $ 6m^2$. But if 3 is not to divide $ k$ as well as $ a+ik$ we must have $ a,d$ of the form $ 3m^2$ or $ 6m^2$, since these differ by a multiple of 3 regardless of $ k$. Let $ a=3\alpha s^2, a+3k=3\beta t^2$ with $ \alpha,\beta = 1 \text{or} 2$. Then $ k=\beta t^2 - \alpha s^2$

Let $ p$ be the common difference of $ a,b,c$ and $ d$. We consider the arithmetic progression in two cases: when $ p$ is odd and when $ p$ is even. Case 1: When $ p$ is odd, Define $ (a_1,b_1,c_1,d_1)$ the residue set of $ (a,b,c,d)$ $ modulo$ $ 4$. Since $ p$ is odd, clearly $ (a_1,b_1,c_1,d_1) = \{a,a + 1,a + 2,a + 3\}$, which is obviously equivalent to the set $ \{0,1,2,3\}$. Therefore, $ abcd$ can be expressed as $ (4m + 2)(4^n)(q)(r) = qr(2^{2n + 1})(2m + 1)$, where $ q$ and $ r$ are positive odd integers and $ m, n$ are positive integers. Hence, $ abcd$ has an odd power of 2, which trivially means $ abcd$ is not a perfect square. Case 2: When $ p$ is even, Note that if $ a$ is even, letting $ k$ be the GCD of $ a$ and $ p$, we can always rewrite $ abcd$ as $ k^4a'(a' + p')(a' + 2p')(a' + 3p')$ $ (a = ka'$ and $ p = kp')$. Obviously, $ gcd(a',p') = 1$ and at least one of them is odd. If $ p'$ is odd, then we can go back to Case 1. When $ a'$ is odd and $ p'$ is even,

Firstly, let us introduce and prove 2 lemmas: Lemma 1. There are no four integers $ m,n,u,v$ with $ \gcd(m,n) = \gcd(u,v) = 1$ and $ mn = uv$ so that $ 9m^2 + n^2 = u^2 + v^2$.

Proof:

Considering $ 9m^2 + n^2 = u^2 + v^2$ modulo 3, we obtain that either $ 3|n,u,v$, which contradicts $ \gcd(u,v) = 1$, or $ 3\not|n$ and 3 divides exactly one of the numbers $ u,v$, say $ u$. Let $ \displaystyle\frac mu = \frac vn = \frac ab$, where $ \gcd(a,b) = 1$. Since $ vb = na$, $ v = ca$ for some integer $ c$. Hence $ n = cb$. Similarly, there is an integer $ d$ so that $ m = da$, $ u = db$. Note that since $ 3\not|v$, we have $ \gcd(3,c) = 1$. Also $ \gcd(u,v) = 1 = \gcd(db,ca)$, hence $ \gcd(c,d) = 1$. Then the relation $ 9m^2 + n^2 = u^2 + v^2$ rewrites as $ 9a^2d^2 + b^2c^2 = a^2c^2 + b^2d^2$, or $ a^2(9d^2 - c^2) = b^2(d^2 - c^2)$. From $ \gcd(a,b) = \gcd(a^2,b^2) = 1$, we deduce that there is a $ k\in\mathbb N_0$, so that $ 9d^2 - c^2 = kb^2$ and $ d^2 - c^2 = ka^2$. Let now $ d - c = x$, $ d + c = y$. Then $ 2d = x + y$, and the last two relations rewrite as $ (2x + y)(2y + x) = kb^2$ and $ xy = ka^2$. Note that $ x,y$ have the same parity. Also, since $ \gcd(c,d) = 1$, $ \gcd(x,y)\le2$. If $ \gcd(x,y) = 2$, then $ 2|x,y$, and $ 4|xy$ and $ 4|(2x + y)(2y + x)$. Since $ a$ or $ b$ must be odd, we get $ 4|k$. Divide then $ x,y$ by 2 and $ k$ by 4, to obtain a completely similar case just with $ x,y$ odd (now $ \gcd(x,y) = 1$). So it is enough to consider the case $ x,y$ odd. Since $ k|xy$ and $ k|(2x + y)(2y + x) = 2(y^2 + x^2) + 5xy$, we have $ k|2(x^2 + y^2)$. Assume $ k > 1$ and let $ p$ be a prime dividing $ k$. Since $ x,y$ are odd and $ k|xy$, $ p > 2$. Then either $ p|x$ or $ p|y$. But then $ k|x^2 + y^2$ implies $ p|x$ and $ p|y$, contradicting $ \gcd(x,y) = 1$. So $ k = 1$. Then $ xy = a^2$ and $ (2x + y)(2y + x) = b^2$. $ \gcd(x,y) = 1$ implies $ x = X^2$ and $ y = Y^2$ for some $ X,Y$. Going back to $ 9d^2 - c^2 = kb^2 = b^2$, we obtain $ \gcd(3,b) = 1$ since $ \gcd(3,c) = 1$. So 3 does not divide $ 2x + y$ or $ 2y + x$. Let $ p$ be a prime divisor of $ \gcd(2x + y,2y + x)$. Then $ p|3(x + y)$ and since $ p\neq3$, $ p|x + y$. Subtracting $ x + y$ from $ 2x + y$ and from $ 2y + x$ we obtain $ p|x,y$ contradicting $ \gcd(x,y) = 1$. Hence $ 2x + y$ and $ 2y + x$ are coprime and their product is a perfect square. Then each of them is a perfect square, say $ x'^2$ and $ y'^2$. But then $ x'^2 + y'^2 = 3(x + y) = 3(X^2 + Y^2)$. The last equation has no solutions in positive intgers since we obtain infinite descent by looking mod 3. Lemma 2. There are no four integers $ m,n,u,v$ with $ n$-odd, $ \gcd(m,n) = \gcd(u,v) = 1$ and $ mn = uv$ so that $ u^2 - v^2 = 8m^2 + n^2$.

Proof:

We proceed similarly to Lemma 1 to obtain $ v = ca$, $ n = cb$, $ m = da$, $ u = db$. Note that $ n$ odd implies $ c$ odd. Then the second equation rewrites as $ 8a^2d^2 + b^2c^2 = d^2b^2 - a^2c^2$, or $ a^2(8d^2 + c^2) = b^2(d^2 - c^2)$. Again $ 8d^2 + c^2 = kb^2$ and $ d^2 - c^2 = ka^2$ for some odd positive integer $ k$. Since $ \gcd(a,b) = 1$, $ 3\not|a^2 + b^2$. Then $ 9d^2 = 8d^2 + c^2 + d^2 - c^2 = k(a^2 + b^2)$ implies $ 9|k$. Let again $ x = d - c$, $ y = d + c$ with $ xy = ka^2$. $ 8d^2 + c^2 = 2(x + y)^2 + \frac14(x - y)^2 = kb^2$, or $ 9(x^2 + y^2) + 14xy = 4kb^2$. Since $ k|xy$, we obtain $ k|9(x^2 + y^2)$. Since $ \gcd(x,y) = 1$, it can be easily shown that $ (xy,x^2 + y^2) = 1$. Then $ k|9$. Together with $ 9|k$, we obtain $ k = 9$. Hence we arrived at equations: $ 8d^2 + c^2 = 9b^2$, $ d^2 - c^2 = 9a^2$. Summing them up, one obtains $ d^2 = a^2 + b^2$. $ \gcd(a,b) = 1$ implies that $ (a,b,d)$ is a pythagorean primitive triple. Note that $ n$ being odd and $ n = cb$ implies $ b$ is odd. Hence there are such positive integers $ f,g$ so that $ b = f^2 - g^2$, $ a = 2fg$, $ d = f^2 + g^2$. Now $ d^2 = c^2 + (3a)^2$, together with $ \gcd(d,c) = 1$ and $ a$ being even implies the existence of such integers $ i,j$ so that $ d = i^2 + j^2$, $ 3a = 2ij$. But then we have $ i^2 + j^2 = f^2 + g^2$ and $ 3fg = ij$. WLOG, $ 3|i$, $ i = 3I$. But then $ 9I^2 + j^2 = f^2 + g^2$ and $ fg = Ij$, with $ \gcd(f,g) = \gcd(I,j) = 1$. However Lemma 1 tells us this system has no non-zero solutions. ___________________________ Without loss of generality, assume $ a < b < c < d$. Let $ r = b - a = c - b = d - c$. Assume that $ abcd = a(a + r)(a + 2r)(a + 3r) = m^2$, for some $ m\in\mathbb Z$. If $ \gcd(a,r) = t > 1$ then let $ a = a't$, $ r = r't$. Then $ a'(a' + r')(a' + 2r')(a' + 3r')t^4 = m^2$, hence $ E = a'(a' + r')(a' + 2r')(a' + 3r')$ is also a perfect square. For simplicity of notations let's work with $ a,r$ instead of $ a',r'$, assuming $ \gcd(a,r) = 1$. So, $ m^2 = [a(a + 3r)][(a + r)(a + 2r)] = (a^2 + 3ar + r^2)^2 - r^4$. So we have the equation $ r^4 + m^2 = (a^2 + 3ar + r^2)^2$. From $ \gcd(a,r) = 1$ we have $ \gcd(r^2,a^2 + 3ar + r^2) = 1$, hence $ (r^2,m,a^2 + 3ar + r^2)$ is a primitive Pythagorean Triple. It is well-known that a primitive Pythagorean triple is of the form $ (u^2 - v^2,2uv,u^2 + v^2)$, where $ \gcd(u,v) = 1$. ___________________________ Case I. $ r$ is even. Then $ a$ is odd. From the above observations, $ r^2 = 2uv$ for some $ \gcd(u,v) = 1$ and $ a^2 + 3ar + r^2 = u^2 + v^2$. From the first relation we infer, WLOG, $ u = 2x^2,v = y^2$ and $ r = 2xy$. Then the second equality becomes $ a^2 + 6xya + 4x^2y^2 = 4x^4 + y^4$, or $ a^2 + 6xya - (2x^2 - y^2)^2 = 0$. Considered as a quadratic equation in $ a$ we must have $ \Delta = 9x^2y^2 + (2x^2 - y^2)^2$ a perfect square. Further, $ \Delta = 4x^4 + y^4 + 5x^2y^2 = (2x^2 + y^2)^2 + x^2y^2 = D^2$, for some integer $ D$. If $ y$ is even then divide the relation by $ 4$ to obtain a similar one with $ x,y$ reversed. Since $ \gcd(x,y) = 1$, $ x$ would be odd, and we would have exactly the original equation. So we can assume $ y$ is odd. It is easy to see that $ \gcd(2x^2 + y^2,xy) = 1$, hence there are such numbers $ m,n$ so that $ xy = 2mn$, $ 2x^2 + y^2 = m^2 - n^2$. Hence $ x$ is even. Let $ x = 2X$. Then $ 8X^2 + y^2 = m^2 - n^2$ for some $ m,n$ coprime and $ Xy = mn$. Then $ mn = Xy$ and $ m^2 - n^2 = 8X^2 + y^2$, which has no solution by Lemma 2 (remember $ y$ is odd, so we fit into its conditions). ___________________________ Case II. $ r$ is odd. Then $ r^2 = u^2 - v^2$ for some $ \gcd(u,v) = 1$. Hence $ u^2 = r^2 + v^2$. Since $ \gcd(u,v) = 1$ we infer that this is another primitive triple, hence there are coprime numbers $ m,n$ so that $ r = m^2 - n^2$ (remember $ r$ is odd), $ v = 2mn$, $ u = m^2 + n^2$. Then the relation $ a^2 + 3ar + r^2 = u^2 + v^2$ rewrites as $ a^2 + 3a(m^2 - n^2) + (m^2 - n^2)^2 = 4m^2n^2 + (m^2 + n^2)^2$, or $ a^2 + 3(m^2 - n^2)a - 8m^2n^2 = 0$. Again, considered as quadratic equation in $ a$, we must have $ \Delta = 9(m^2 - n^2)^2 + 32m^2n^2$ a perfect square. Further, we obtain $ \Delta = 9(m^2 + n^2)^2 - 4m^2n^2 = D^2$ for some integer $ D$. This rewrites as another Pythagorean Equation $ (2mn)^2 + D^2 = (3m^2 + 3n^2)^2$. Hence $ 3|2mn$ and $ 3|D$. Since $ \gcd(m,n) = 1$, $ 3$ divides exactly one of the numbers $ m,n$, say $ m$. Let $ m = 3M$ and $ D = 3D'$. Then $ (2Mn)^2 + D'^2 = (9M^2 + n^2)^2$. Since $ \gcd(m,n) = 1$, we easily get $ \gcd(2Mn,9M^2 + n^2) = 1$, hence $ (2Mn,D',9M^2 + n^2)$ is a primitive Pythagorean Triple, so $ 2Mn = 2PQ$, $ 9M^2 + n^2 = P^2 + Q^2$ for some coprime integers $ P,Q$. However, by Lemma 1, there are no such integers $ M,n,P,Q$.

If $ a,b,c,d$ are in an arithmetic progression, it means that one can rewrite $ (a,b,c,d)$ as $ (x-3k,x-k,x+k,x+3k)$ being $ k\not= 0$. $ (x-3k)(x-k)(x+k)(x+3k)= (x^2-k^2)(x^2-9k^2)=x^4-10k^2x^2+9k^4$ Substitute $ x^2=y$ then considere $ P(y)=y^2-10k^2y+9k^4$ We know that it is a square iff $ \Delta=0$ $ \Delta=4(25k^4-9k^4)=4(16k^4)$ which can't be $ 0$ because it contradicts the hypothesis $ k\not= 0$.

Let $ b=a+r, c=a+2r, d=a+3r$. It follows that $ a(a+r)(a+2r)(a+3r)=p^2 \Rightarrow (a^2+3ar)(a^2+3ar+2r^2)=p^2$ If $ gcd(a,r)=d \not= 1$ then $ a=a_1d$ and $ r=r_1d$, where $ gcd(a_1,r_1)=1$ $ p^2=d^4(a_1^2+3a_1r_1)(a_1^2+3a_1r_1+2r_1^2) \Rightarrow$ $ (a_1^2+3a_1r_1)(a_1^2+3a_1r_1+2r_1^2)$ is a perfect square. So, we can suppose that $ gcd(a,r)=1$. 1. Suppose that there exist a prime $ q$ such that $ q|gcd(a^2+3ar,a^2+3ar+2r^2)$. We conclude that: $ q|(a^2+3ar+2r^2)-(a^2+3ar)=2r^2$ If $ q|r^2 \Rightarrow q|r \Rightarrow q|3ar \Rightarrow q|(a^2+3ar)-3ar=a^2 \Rightarrow q|a \Rightarrow gcd(a,r) \not= 1$, contradiction. It follows that $ q=2$. It is easy to see that if $ 4|a^2+3ar$ and $ 4|a^2+3ar+2r^2 \Rightarrow 4|2r^2 \Rightarrow 2|r \Rightarrow 2|a^2 \Rightarrow 2|a \Rightarrow gcd(a,r) \not= 1$, contradiction. In this case, $ gcd(a^2+3ar,a^2+3ar+2r^2)=2 \Rightarrow \exists x,y \in \math{N}^*, gcd(x,y)=1$, such that $ a^2+3ar=2x^2$ $ a^2+3ar+2r^2=2y^2$ $ a^2+3ar+2r^2=2y^2 \Rightarrow (a+r)(a+2r)=2y^2$ If $ \exists c>1$, such that $ c|a+r, c|a+2r \Rightarrow c|r \Rightarrow c|a \Rightarrow gcd(a,r)\not= 1$, contradiction. We conclude that $ gcd(a+r,a+2r)=1 \Rightarrow$ $ a+r=m^2$ $ a+2r=n^2$ So, the exponent of $ 2$ in $ (a+r)(a+2r)$ is even, but the exponent of $ 2$ in $ 2y^2$ is odd, contradiction. 2. $ gcd(a^2+3ar,a^2+3ar+2r^2)=1 \Rightarrow$ $ a^2+3ar=x^2$ $ a^2+3ar+2r^2=y^2$ and $ gcd(x,y)=1$ $ \Rightarrow 2r^2+x^2=y^2 \Rightarrow 2r^2=y^2-x^2=(y-x)(y+x)$ If $ \exists k \in \math{N}^*$ such that $ k|y-x$ and $ k|y+x \Rightarrow k|2y, k|2x \Rightarrow k|gcd(2x,2y)=2gcd(x,y)=2$ If $ k=1$ then $ gcd(y-x,y+x)=1$. Since $ y-x,y+x$ have the same parity $ \Rightarrow y-x$ is odd and $ y+x$ is odd $ \Rightarrow (y-x)(y+x)$ is odd $ \Rightarrow 2r^2$ is odd, contradiction. It follows that $ k=2 \Rightarrow gcd(y-x,y+x)=2 \Rightarrow y-x=2(y_1-x_1)$ and $ y+x=2(y_1+x_1)$, where $ gcd(y_1-x_1,y_1+x_1)=1 \Rightarrow 4(y_1-x_1)(y_1+x_1)=2r^2 \Rightarrow (y_1-x_1)(y_1+x_1)=2r_1^2$ We obtain the same contradiction. Since $ gcd(y_1-x_1,y_1+x_1)=1$ and $ y_1-x_1,y_1+x_1$ have the same parity $ \Rightarrow (y_1-x_1)(y_1+x_1)$ is odd, but $ 2r_1^2$ is even, contradiction.

Let $ a$ be the starting member of the progression, and $ b$ its difference. We suppose $ a(a + b)(a + 2b)(a + 3b) = x^2$ for some $ x$. We can WLOG assume $ (a,b) = 1$ because if not we cant divide the equality with it ... If we transform the expression a little bit we get: $ (a^2 + 3ab + b^2)^2 = (b^2)^2 + x^2$ so we get a pythagorean triplet $ (a,b) = 1$: $ a^2 + 3ab + b^2 = m^2 + n^2$ $ b^2 = m^2 - n^2$ $ x = 2mn$ , where $ (m,n) = 1$ So, we get: $ a(a + 3b) = 2n^2$ and $ (a + b)(a + 2b) = 2m^2$ , so $ b$ is odd, beacause if its even, we get $ (a + b,a + 2b) \geq 2$ , which is impossible because $ (a,b) = 1$. Now, if $ 2|a$ then $ a = 2p^2, a + 3b = q^2, a + 2b = 2r^2, a + b = t^2$. Because $ a + 3b$, $ a + b$ are odd and $ (a + 3b) - (a + b) = 2b \equiv2 mod 4$ then one of $ t^2,q^2 \equiv3 mod 4$ which is a contradiction. If $ a$ is odd, then we have $ a \equiv1 mod 4$ and then we get $ a + 2b = r^2 \equiv3 mod 4$ which is a contradiction again.

- assume the statement wrong. than there exist such $ a,b,c,d$ that $ abcd$ is a perfect square - we can denote $ b = a + n$, $ c = a + 2n$ and $ d = a + 3n$ where $ a$ is a positie integer and $ n$ an integer different to zero. so $ abcd = a(a + n)(a + 2n)(a + 3n)$ is a perfect square. WLOG assume $ gcd(a,n) = 1$ (if not we can just divide the product $ abcd$ by the fourth power of gratest common devisor of $ a$ and $ n$) - there exist no 4perfect squares in arithmetical progression. this is a known fact. 1) - if $ n$ is even than $ a$ must be odd (they are relatively prime). than $ a + n$ and $ a + 2n$ are relatively prime to all $ a,b,c,d$ therefore due to lemma1 $ gcd(a,a + 3n) = 3$ therefore we denote $ 2l = n$ and $ 3k = a + 3l$ and we have $ a(a + n)(a + 2n)(a + 3n) = 9(k - l)(3k - l)(3k + l)(k + l)$ menaing each of $ (k - l),(3k - l),(3k + l),(k + l)$ is a perfect square.(since they are now all relatively prime to each other and their product is a perfect square) - $ k^2 - l^2$ is therefore also a perfect square. meaning $ k$ is odd (due to pythagorean triples formula) ,meaning $ l$ is even since $ 3k + l = a + 2n$ is odd. - on the other hand $ (3k + l)(k + l) = 3k^2 + 4kl + l^2 = p^2$ is a perfect square. now we have $ p^2 + k^2 = (2k + l)^2$ which is a primitive pythagorean triple. meaning $ 2k + l$ is odd leading to $ l$ odd - such $ l$ that is odd and even doesnt exist... therefore $ n$ is odd 2) - if $ n$ is odd $ a$ and $ a + 3n$ are of different parity. we can assume $ a$ is even (because of the simmerty due to the deffinition of $ a$ and $ n$, otherwise denote $ a + 3n = a',a + 2n = a' + ( - n)$ etc.) - this means $ a + n$ is a perfect square. - if $ a(a + 2n) = a^2 + 2an = q^2$ was a perfect square (than so would be $ a + 3n$). than we would have $ q^2 + n^2 = (a + n)^2$ which is a primitive pythagorean triple. menaing $ n = u^2 - v^2$ and $ a + n = u^2 + v^2$(for some relatively prime $ u,v$ of differnent parity). meaning $ a + 3n = 3u^2 - v^2$ which obviously has no solutions modulo $ 4$ due to definition of $ u,v$ and due to the fact that $ a + 3n$ is a perfect square. therefore $ a(a + 2n)$ is not a perfect square but $ a(a + 2n)(a + 3n)$ has to be - denote $ a = 6r$ leading to $ r(6r + n)(3r + n)(2r + n)$ is a perfect square (where all of $ r,6r + n,3r + n,2r + n$ are relatively prime to each other and each of them is a perfect square) - $ r+(2r+n)=(3r+n)$ is therefore a primitive pythagorean triple meaning $ r=(2xy)^2$

Let rewrite the problem as follow: Prove that $ a(a + d)(a + 2d)(a + 3d) = m^2$ do not have positive integral solution. We can assume $ (a,d) = 1$, otherwise we can divide both sides by the common factor. If there exists a solution, then $ (a^2 + 3ad)(a^2 + 3ad + 2d^2) = m^2$ Since $ (a^2 + 3ad,a^2 + 3ad + 2d^2) = (a^2 + 3ad,2d^2) = (a^2 + 3ad,2)$ $ (a^2 + 3ad,a^2 + 3ad + 2d^2) = 1$ or $ 2$ If $ d$ is even, then $ a$ is odd and $ a^2 + 3ad$ is also odd. Thus $ (a^2 + 3ad, a^2 + 3ad + 2d^2) = 1$, so $ a^2 + 3ad = p^2$, $ a^2 + 3ad + 2d^2 = q^2$, with $ (p,q) = 1$ and $ pq = m$ $ (a + d)(a + 2d) = q^2$ and $ (a + d,a + 2d) = (a + d,d) = 1$ Therefore $ a + d = x^2$, $ a + 2d = y^2$, with $ (x,y) = 1$ and $ xy = q$ Now $ a = 2x^2 - y^2$, $ d = y^2 - x^2$ and $ a + 3d = 2y^2 - x^2$ Thus $ (2x^2 - y^2)(2y^2 - x^2) = p^2$ If $ (2x^2 - y^2,2y^2 - x^2) = c$, and $ c > 1$, then let $ k$ be a prime factor of $ c$ Then $ x^2 + y^2 = (2x^2 - y^2) + (2y^2 - x^2)$ and $ 3(y^2 - x^2) = (2y^2 - x^2) - (2x^2 - y^2)$ are divisible by $ k$ If $ k = 3$, then $ 2y^2 - x^2$ is multiple of 3, which is only possible when $ x,y$ are multiple of 3. Contradicts with $ (x,y) = 1$ If $ k$ is a factor of $ x^2 - y^2$, then $ 2x^2$ and $ 2y^2$ are multiple of $ k$. If $ k = 2$, $ 2x^2 - y^2$ is even, and $ y$ is even. Similarly $ x$ is even, also a contradiction. Then $ k$ is factor of $ x^2$ and $ y^2$, which is also impossible. Therefore $ (2x^2 - y^2,2y^2 - x^2) = 1$ and $ 2y^2 - x^2$ and $ 2x^2 - y^2$ are both perfect square. Then $ 2y^2 - x^2$, $ y^2$, $ x^2$, $ 2x^2 - y^2$ are four perfect square in an arithmetic sequence, which is impossible. Therefore $ d$ must not be even. If $ d$ and $ a$ are both odd, $ (a^2 + 3ad,a^2 + 3ad + 2d^2) = 2$ so $ a^2 + 3ad = 2p^2$, $ a^2 + 3ad + 2d^2 = 2q^2$, with $ (p,q) = 1$ and $ 2pq = m$ $ (a + d)(a + 2d) = 2q^2$ and $ (a + d,a + 2d) = (a + d,d) = 1$ Therefore $ a + d = 2x^2$, $ a + 2d = y^2$, with $ (x,y) = 1$ and $ xy = q$ Then $ d = y^2 - 2x^2$, $ a = 4x^2 - y^2$, $ a + 3d = 2(y^2 - x^2)$ Thus $ (y^2 - x^2)(4x^2 - y^2) = p^2$ $ (y^2 - x^2,4x^2 - y^2) = (y^2 - x^2,3x^2) = (y^2 - x^2,3) = 1$ or $ 3$ If $ y^2 - x^2$ is a multiple of $ 3$, then $ y^2 - x^2 = 3g^2$, $ 4x^2 - y^2 = 3h^2$, with $ (g,h) = 1$ and $ 3gh = p$ Then $ g^2 + h^2 = x^2$, and $ 4g^2 + h^2 = y^2$ We will show that $ g^2 + h^2$ and $ 4g^2 + h^2$ cannot be both perfect square. By $ d=y^2-2x^2$, we find that $ g$ is even and $ h$ is odd. Since $ g^2+h^2=x^2$, $ g=2ef$, $ h=e^2-f^2$ where $ (e,f)=1$ Since $ 4g^2+h^2=y^2$, $ g=kl$, $ h=k^2-l^2$ where $ (k,l)=1$ Then $ 2ef=kl$, there exists $ A,B,C,D$ such that they are co-prime to each other, and $ e=AB,f=CD,k=2AC,l=BD$ or $ e=AB,f=CD,k=AC,l=2BD$ In the first case, by $ e^2-f^2=k^2-l^2$, we have $ B^2(A^2+D^2)=C^2(4A^2+D^2)$ Since $ (B^2,C^2)=(A^2+D^2,4A^2+D^2)=1$, $ A^2+D^2$ and $ 4A^2+D^2$ are both perfect square but $ AD<ef<gh$, using infinite descent, we can prove the statement. The second case is similar. So $ (y^2 - x^2,4x^2 - y^2) = 1$, and then $ y^2 - x^2 = i^2$, $ 4x^2 - y^2 = j^2$ with $ (i,j) = 1$ and $ ij = p$ Then $ 3x^2 = i^2 + j^2$, $ i$ and $ j$ must be multiple of 3, impossible. If $ d$ is odd and $ a$ is even, $ (a^2 + 3ad,a^2 + 3ad + 2d^2) = 2$ so $ a^2 + 3ad = 2p^2$, $ a^2 + 3ad + 2d^2 = 2q^2$, with $ (p,q) = 1$ and $ 2pq = m$ $ (a + d)(a + 2d) = 2q^2$ and $ (a + d,a + 2d) = (a + d,d) = 1$ Therefore $ a + d = x^2$, $ a + 2d = 2y^2$, with $ (x,y) = 1$ and $ xy = q$ Then $ d = 2y^2 - x^2$, $ a = 2x^2 - 2y^2$, $ a + 3d = 4y^2 - x^2$ Thus $ (x^2 - y^2)(4y^2 - x^2) = p^2$ Just like the argument in the above case, this is impossible. Therefore, in conclusion, this equation has no positive integral solution.

Assume that $ a(a+d)(a+2d)(a+3d)$ is a perfect square for some minimal positive integers $ a,d$. Then clearly $ (a,d)=1$. Also note that $ S=a(a+d)(a+2d)(a+3d)=(a^2+3ad+d^2)^2-d^4$. Therefore if $ S$ is a perfect square we must have $ a^2+3ad+d^2=x^2+y^2$ and $ d^2\in\{2xy, x^2-y^2\}$ for some relatively prime positive integers $ x,y$ such that $ x+y$ is odd. Case 1: $ d^2=2xy$ This implies $ a^2+3ad=a(a+3d)=(x-y)^2$. Verify that $ (a,a+3d)\mid 3$ since $ (a,d)=1$. First assume that $ (a,a+3d)=1$. Then $ 3\not|\,a$ and $ a=u^2, a+3d=v^2\implies d=\frac{v^2-u^2}{3}$ for some positive integers $ u<v$ such that $ (u,v)=1$. Hence $ S=a(a+d)(a+2d)(a+3d)$ implies $ (a+d)(a+2d)$ is a perfect square. But $ (a+d)(a+2d)=\left(\frac{v^2+2u^2}{3}\right)\left(\frac{2v^2+u^2}{3}\right)$. Also note that $ g=\left(\frac{v^2+2u^2}{3},\frac{2v^2+u^2}{3}\right)\mid u^2\wedge v^2\implies g=1$. Thus $ v^2+2u^2=3w_1^2$ and $ 2v^2+u^2=3w_2^2$, for some integers $ w_1<w_2$. Therefore we get $ u^2=2w_1^2-w_2^2$ and $ v^2=2w_2^2-w_1^2$. Note that $ u<v\implies w_2>w_1$ and $ u<w_1, v>w_2$. From the former equation we have $ u^2+w_2^2=2w_1^2\Leftrightarrow \left(\frac{w_2+u}{2}\right)^2+\left(\frac{w_2-u}{2}\right)^2=w_1^2$. Hence $ \frac{w_2\pm u}{2}=s^2-t^2$ and $ \frac{w_2\mp u}{2}=2st$ and $ w_1=s^2+t^2$ (for some positive integers $ s>t$) which together imply $ w_2=s^2-t^2+2st$ and $ w_1=s^2+t^2$ and $ u=\pm (s^2-t^2-2st)$. Similarly from the latter equation we get $ w_2=s'^2+t'^2, v=s'^2-t'^2+2s't'$ and $ w_1=\pm(s'^2-t'^2-2s't')$ for some positive integers $ s'>t'$ (remember $ v>w_2$). To be continued later...

definations we call a quadruple of numbers a staton if their product is a perfect square $ (a,d)$ denotes the gcd of $ a$ and $ d$ $ ap$ stands for arithematic progression _________________________________________ some obvous lemmas used not mentioned 1 if $ mn$ $ =$ $ p^{2}$ and $ (m,n) = 1$ then $ m$ and $ n$ are perfect squares 2 if $ m^{2} + n^{2} = p^{2}$ then there exist $ u$ and $ v$ such that $ u^{2} + v^{2} = p$ and either {$ u^{2} - v^{2} = n$ and $ 2uv = m$ } or {$ u^{2} - v^{2} = m$ and $ 2uv = n$ } 3 odd squares are congruent to $ 1$ mod $ 8$ 4 squares are congruent to $ 0$ or $ 1$ mod $ 3$ 5 fermats theorem that says no $ 4$ squares are in an $ ap$ proof: proof goes with infinite desent over the common difference(which we may assume positive)i will prove the theorem for the common difference even and all terms in the $ ap$ coprime to each other (for i have proved that the ap considered can be assumed to satisfy these properties ) let the $ ap$ be $ x - 3n,x - n,x + n,x + 3n$ if two squares differ by an even integer $ 2n$ they must differ by an integer divisible by $ 4$; so $ n$ must be even and $ x$ is odd. as they themselves are squares their product will also be a square therefore there exist $ y$ such that $ y^{2} = x^{4} - 10x^{2}n^{2} + 9n^{2} = (x^{2} - 5n^{2})^{2} - 16n^{4}$ therefore there exist $ 2$ relatively co prime integers $ u$ and $ v$ such that $ 4uv = 4n^{2}$ and $ (x^{2} - 5n^{2})^{2} = 4u^{2} + v^{2}$ from the first equation it follows that u and v are perfect squares as $ (u,v) = 1$ let $ u = A^{2}$ and $ v = D^{2}$ therefore $ x^{2} = (4A^{2} + D^{2})(A^{2} + D^{2})$ and $ n = AD$ as $ (A,D) = 1$, $ (4A^{2} + D^{2},A^{2} + D^{2}) = 1$ and thus $ 4A^{2} + D^{2}$ and $ A^{2} + D^{2}$ are perfect squares applying lemma $ 2$ repetatively we get that there exist $ q,r,s,t$ pairwise relatively coprime and $ q$ even such that $ 2A = 2 \cdot 2qr \cdot st = 4qs \cdot rt$ and $ s^{2}t^{2}$ $ -$ $ q^{2}r^{2} = r^{2}t^{2}$ $ -$ $ 4q^{2}s^{2} = D$ thus $ (4q^{2} + t^{2})s^{2} = r^{2}(q^{2} + t^{2})$ multiplying both sides by $ q^{2} + t^{2}$ we get $ (4q^{2} + t^{2})(q^{2} + t^{2}) = k^{2}$ for some $ k$ (which can be evaluated ) and as $ 2qt$ divides $ A$ ,$ qt$ $ <$ $ A$ $ \le$ $ AD$ thus we have an infinitly decreasing sequence of positive integers which is a contradiction unless $ AD = n = 0$ _________________________________________ outlne 1 $ a$ is odd (this was the most tedous ) 2 $ a$ is not divisible by $ 3$ (trivial ) 3 all terms in the ap are perfect squares (elegance ) 4 use fermats thm that states no $ 4$ squares are in an $ ap$ (elegance ) _________________________________________ solution let $ {a, a + d, a + 2d, a + 3d}$ be a staton and $ d$ not equal to $ 0$ then obviously $ {a/(a,d),a/(a,d) + d/(a,d),a/(a,d) + 2d/(a,d),a/(a,d) + 3d/(a,d)}$ also is a staton therefore we take the liberty to assume $ (a,d) = 1$ let their product $ = k^{2}$ we get $ (a^{2} - 5 \cdot d^{2})^{2} = d^4 + k^{2}$ if $ d$ is odd then considering this as a pythagorean triplet we know that there exist $ x$ and $ y$ such that $ x^{2} - y^{2} = d^{2}$ and $ x^{2} + y^{2} = a^{2} - 5 \cdot d^{2}$ this implies $ 2 \cdot y^{2} + 6 \cdot d^{2} = a^{2}$ consider a solution to the equation above where $ a$ is minimun we observe that $ a$ and $ y$ have to be divisible by $ 3$ and thus obtain a solution $ a/3$,$ y/3$ and $ d/3$ therefore the solution set has no minimal element and thus no solution this equation has no solution therefore $ d$ is even and $ a$ is odd as $ (a,d) = 1$ this tells us $ (a + d,a) = (a + d,a + 2d) = (a + d,a + 3d) = 1$ this implies $ a + d$ is a perfect square also $ (a + 2d,a) = (a + 2d,a + d) = (a + 2d,a + 3d) = 1$ this implies $ a + 2d$ is a perfect square this also implies $ a + d,a + 2d$ are congruent to $ 1$ or $ 0$ mod $ 3$ checking these congruences we deduce that $ a$ is not divisible by $ 3$ therefore (using the euclids algorithm ofcource) we get $ (a,a + 2d) = (a,a + d) = (a,a + 3d) = 1$ this implies that $ a$ is a perfect square also $ (a + 3d,a + 2d) = (a + 3d,a + d) = (a + 3d,a) = 1$ this implies $ a + 3d$ is a perfect square $ {a, a + d, a + 2d, a + 3d}$ are perfect squares this is a contradiction as fermat's theorom of squares in an $ ap$ which states that there cannot exist $ 4$ squares in an $ ap$

Let's suppose that the numbers are $ x,x + y,x + 2y,x + 3y$ and their product is a perfect square. Let $ gcd(x,y) = d$ and then $ x = dx_0$ and $ y = dy_0$ . We observe that the product of numbers $ x_0,x_0 + y_0,x_0 + 2y_0,x_0 + 3y_0$ are also perfect square so we can suppose that $ gcd(x,y) = 1$ Now we have $ gcd(x,x + y) = 1$ ,$ gcd(x + y,x + 2y) = 1$ , $ gcd(x + 2y,x + 3y) = 1$, $ gcd(x + y,x + 2y) = 1\ or\ 2$ , $ gcd(x + y,x + 3y) = 1\ or \ 2$ and $ gcd(x + y,x + 3y) = 1\ or \ 3$ and also $ gcd(x + 2y,x)$ and $ gcd(x + y,x + 3y)$ cannot be both 2. So we have in all 6 cases to consider . $ 1.\ x = a^2,\ x + y = b^2,\ x + 2y = c^2,\ x + 3y = d^2$ $ 2.\ x = a^2,\ x + y = 2b^2,\ x + 2y = c^2,\ x + 3y = 2d^2$ $ 3.\ x = 2a^2,\ x + y = b^2,\ x + 2y = 2c^2,\ x + 3y = d^2$ $ 4.\ x = 3a^2,\ x + y = b^2,\ x + 2y = c^2,\ x + 3y = 3d^2$ $ 5.\ x = 3a^2,\ x + y = 2b^2,\ x + 2y = c^2,\ x + 3y = 6d^2$ $ 6.\ x = 6a^2,\ x + y = b^2,\ x + 2y = 2c^2,\ x + 3y = 3d^2$ where in all cases $ a,b,c,d$ are paiwise relatively prime numbers. In the first case we have to prove that four consecutive terms in an arithmetic sequence cannot all be squares, suppose there exist four squares $ A^2, B^2, C^2, D^2$ in increasing arithmetic progression, i.e., we have $ B^2 - A^2 = C^2 - B^2 = D^2 - C^2$. We can assume the squares are mutually co-prime, and the parity of the equation shows that each square must be odd. Hence we have co-prime integers u,v such that $ A = u - v, C = u + v, u^2 + v^2 = B^2$, and the common difference of the progression is $ \frac {C^2 - A^2}{2} = 2uv$. We also have $ D^2 - B^2 = 4uv$, which factors as $ (\frac {D + B}{2})(\frac {D - B}{2}) = uv$. The two factors on the left are co-prime, as are u and v, so there exist four mutually co-prime integers a,b,c,d (exactly one even) such that $ u = ab, v = cd, D + B = 2ac$, and $ D - B = 2bd$. This implies $ B = ac - bd$, so we can substitute into the equation $ u^2 + v^2 = B^2$ to give $ (ab)^2 + (cd)^2 = (ac - bd)^2$. This quadratic is symmetrical in the four variables, so we can assume c is even and a, b, d are odd. From this quadratic equation we find that c is a rational function of the square root of $ a^4 - a^2d^2 + d^4$ , which implies there is an odd integer m such that $ a^2 - a^2d^2 + d^4 = m^2$. Since a and d are odd there exist co-prime integers x and y such that $ a^2 = k(x + y)$ and $ d^2 = k(x - y)$, where $ k = 1$ or $ k = - 1$. Substituting into the above quartic gives $ x^2 + 3y^2 = m^2$, from which it's clear that y must be even and x odd. Changing the sign of x if necessary to make m+x divisible by 3, we have $ 3(\frac {y}{2})^2 = (\frac {m + x}{2})(\frac {m - x}{2})$, which implies that $ \frac {m + x}{2}$ is three times a square, and $ \frac {m - x}{2}$ is a square. Thus we have co-prime integers r and s (one odd and one even) such that $ \frac {m + x}{2} = 3r^2$, $ \frac {m - x}{2} = s^2$ , $ m = 3r^2 + s^2$ , $ x = 3r^2 - s^2$, and $ y = 2rs$ or $ y = - 2rs$. Substituting for x and y back into the expressions for $ a^2,\ d^2$ (and transposing if necessary) gives $ a^2 = k(s + r)(s - 3r)$ and $ d^2 = k(s - r)(s + 3r)$. Since the right hand factors are co-prime, it follows that the four quantities $ (s - 3r), (s - r), (s + r), (s + 3r)$ must each have square absolute values, with a common difference of 2r. These quantities must all have the same sign, because otherwise the sum of two odd squares would equal the difference of two odd squares, i.e., 1+1 = 1-1 (mod 4), which is false. Therefore, we must have $ 3r < s$, so from $ m = 3r^2 + s^2$ we have $ 12r^2 < m$. Also the quartic equation implies $ m < a^2 + d^2$ . Thus we have four squares in arithmetic progression with the common difference $ 2r < 2abcd$, the latter being the common difference of the original four squares. This contradicts the fact that there must be a smallest absolute common difference for four squares in arithmetic progression, so the proof is complete In the second case we have that $ 6b^2 - 2a^2 = 2d^2$ or $ 3b^2 = a^2 + d^2$ , from which we conclude that $ 3|a$ and $ 3|d$ , a contradiction In the third case we have $ 3b^2 - 4a^2 = d^2$ and considering mod 3 we have that $ 3|a$ and $ 3|d$ , a contradiction . In the forth case we have that $ 2b^2 - 3a^2 = c^2$ and takind mod 3 we have that $ 3|b^2 + c^2$ so $ 3|b$ and $ 3|c$ ,again a contradiction . Now since the cases (5) and (6) are similar by reversing the order of {a,b,c,d} we will just treat one of them , say the 5. In this case we have that $ a^2 = b^2 - d^2$ or $ a^2 = (b - d)(b + d)$ which gives $ b + d = r^2$ ,$ b - d = s^2$ or $ b + d = 2r^2$ , $ b - d = 2s^2$ But we have also that $ c^2 = b^2 + 3d^2$ so subtituting the previous relations to this we have that $ (\frac {c}{2})^2 = r^4 - r^2s^2 + s^4$ or (if the second case holds) $ c^2 = 2(r^4 - r^2s^2 + s^4)$ In both cases we have no solutions since r and s cannot be both even so $ c^2$ is divisible by $ 2$ and not by $ 4$ , a contradiction again . So we are done

$ Let(t_1;t_2)$ be the LCM of $ t_1$ and$ t_2 ; (t_1;t_2;t_2)$ be the LCM of $ t_1;t_2;t_3$. Lemma1: The possitive integer roots of the equation $ x^2 + y^2 = z^2$ (with x is an odd integer, y is an even integer) are $ x = m^2 - n^2; y = 2mn; z = m^2 + n^2$ (with(m;n)=1; m>n; if m is odd, n will be even; if n is odd, m will be even) Proof of lemma 1: This is the Phytagore equation. Lemma2: The positive integer roots of the equation $ x^2 + y^2 = 2z^2$(with pair of them is the relatively prime integer) are $ x = (m + n)^2 - 2n^2; y = (m + n)^2 - 2m^2; z = m^2 + n^2$ (with (m;n)=1; m>n; if m is odd, n will be even; if n is odd, m will be even). Proof of the lemma2: It is easy to prove that x, y, z be odd integer. Denote by $ u = \frac {x + y}{2}; v = \frac {x - y}{2}$, we have (x;y)=1 and $ 2(x^2 + y^2) = 4x^2$. Consequently $ (\frac {x + y}{2})^2 + (\frac {x - y}{2})^2$.Thus $ u^2 + v^2 = z^2$. Applying the lemma1 to u, v, z. We obtain $ v = 2mn; u = m^2 - n^2; v = m^2 + n^2$.Hence, we obtain $ x = (m + n)^2 - 2n^2; y = (m + n)^2 - 2m^2; z = m^2 + n^2$. Lemma3: the equation $ x^2 + y^2 = 3z^2$ doesn't have possitive integer roots. Proof of the lemma3: We have to prove that if this equation have positive integer root : $ x_0,y_0,z_0$, it will have positive integer root $ \frac {x_0}{3},\frac {x_0}{3},\frac {x_0}{3}$.It is easy and that is a contracdiction. Lemma4:The equation $ 3x^2 + y^2 = 2z^2$ doesn't have positive integer roots. Proof of the lemma4: Similarly (Lemma3). SOLUTION: Suppose without loss of generality that a<b<c<d and denote by a; b=a+k; c=a+2k; d=a+3k (k$ \in$N). Assume that $ abcd = s^2$ (it is an perfect square). Case1: If pair of a,b,c,d are relative prime integer, we will have $ a = x^2, b = y^2, c = z^2, d = t^2$ (y, y, z, t $ \in$ N).Then we have $ x^2 + z^2 = 2y^2 ; y^2 + t^2 = 2z^2$ Applying the lemma2, there exists m,n,m',n' satisfying: $ x = (m + n)^2 - 2n^2; z = (m + n)^2 - 2m^2; y = m^2 + n^2$ And $ y = (m' + n')^2 - 2n'^2; t = (m' + n')^2 - 2m'^2; z = m'^2 + n'^2$. On other hand, $ z^2 - x^2 = t^2 - y^2$. Consequently $ mn(m^2 - n^2) = m'n'(m'^2 - n'^2)$. (1) And $ y^2 - z^2 = z^2 - t^2$.Consequently $ 2m^2n^2 - 4mn(n^2 - m^2) = 2m'^2n'^2 - 4m'n'(m'^2 - n'^2)$. (2) (1), (2) give us $ m^2n^2 = m'^2n'^2; m^2 - n^2 = m'^2 - n'^2$. Thus m=m'; n=n'. That is a contradiction. Case2: Existing two numbers have LCM greater than 1. *If (a;a+k)=r$ \not=$1. We have a, k $ \equiv$0 (mod r). Deducing a, b, c, d $ \equiv$0 (mod r). We consider this problem with $ a' = \frac {a}{r},b' = \frac {b}{r},c' = \frac{c}{r},d' = \frac {d}{r}$..... Similarly, if (a+k;a+2k)$ \not =$1, (a+2k;a+3k)$ \not =$1..... *If (a;a+2k)=r$ \not=$1. Let r' be the prime commonmutiple of a and a+2k. Therefore 2k$ \equiv$0(mod r'). If r'$ \not =$2, we will have a,k $ \equiv$0(modr').We consider this problem with $ a' = \frac {a}{r}, b' = \frac {b}{r}, c' = \frac {c}{r},d' = \frac {d}{r}$...... And summaryly, we have to prove this problem with r=2. Analogously we consider (a+k;a+3k).But (a;a+2k)=2, we have (a+k;a+3k)=1 (because if (a;a+2k)=2, we will consider this problem with $ a' = \frac {a}{2}, b' = \frac {b}{2}, c' = \frac {c}{2},d' = \frac {d}{2}$) *Analogously, we consider (a;a+3k)=w. And we have to consider w=1; w=3. i) if w=1 , now we can show $ a = 2p^2_1; b = p^2_2; c = 2p^2_3; d = p^2_4$. Because a, b, c, d are four distinct positive integers in arithmetic progression, we have: $ p_2^2 = p_1^2 + p_3^2; 4p_3^2 = p_2^2 + p_4^2$. Consequently, $ 3p_3^2 = p_1^2 + p_4^2$. According to the lemma3, this equation doesn't have positive integer. ii)if w=3, we can show $ a = 6p^2_1; b = p^2_2; c = 2p^2_3; d = 3p^2_4$ (pair of $ p_1, p_2, p_3, p_4$ is relatively prime integer). And we have $ p_2^2 = p_3^2 + 3p_1^2; 4p_3^2 = p_2^2 + 3p_4^2$.It follows that $ p_2^2 = p_4^2 + p_1^2; p_3^2 = p_4^2 + 4p_1^2$. Since the lemma1, there exists m, n, m', n' (with (m,n)=1; (m',n')=1; m>n; m'>n'; if m is odd, n will be even; if n is odd, m will be even; if m' is odd, n' will be even; if n' is odd, m' will be even) satisfying: $ 2p_1 = 2mn; p_4 = m^2 - n^2; p_2 = m^2 + n^2$ and $ p_1 = 2m'n'; p_4 = m'^2 - n'^2; p_3 = m'^2 + n'^2$. Hence, we have $ mn = 2m'n'$; $ m^2 - n^2 = m'^2 - n'^2$. It follows that $ m^2 - n^2 = (\frac {mn}{2n'})^2 - n'^2$. Consequently $ n'^2 = \frac { - 4(m^2 - n^2) + 2 \sqrt {m^4 - m^2n^2 + n^4}}{4}$=$ -m^2+n^2 + \frac{\sqrt {m^4 - m^2n^2 + n^4}}{2}$. But if m is odd, n will be even; if n is odd, m will be even. Therefore $ n'^2$ is non-integer.That is a contradiction. *And if (a;a+2k)=1; (a;a+3k)=3, we can show $ a = 3p^2_1; b = p^2_2; c = p^2_3; d = 3p^2_4$. Because a, b, c, d are four distinct positive integers in arithmatic progression, we have: $ 2p_2^2 = 3p_1^2 + p_3^2; 2p_3^2 = 3p_4^2 + p_2^2$.According to the lemma4, this equation doesn't have positive integer roots. THE SOLUTON IS COMPLETE.

If a,b,c,d satisfies the condition, then so did 2a,2b,2c,2d. Therefore it is enough to consider a case when b-a is even. In this case, we can assume that $ (a,b,c,d) = (x - 3y, x - y, x + y, x + 3y)$ for some natural number $ x,y$. Then $ abcd = (x^{2} - y^{2})(x^{2} - 9y^{2}) = (x^{2} - 5y^{2})^{2} - 16y^{2}$. To be a square number, $ abcd \le (x^{2} - 5y^{2} - 2)^{2}$. ( $ abcd \equiv x^{2} - 5y^{2} \pmod2$) So $ 4(x^{2} - 5y^{2}) - 4 \le 16y^{2}$ But we know that $ x - 3y = a$ greater than 0. So we get $ 16y^{2} \ge 4((3y + 1)^{2} - 5y^{2}) - 4 = 4(4y^{2} + 6y + 1) - 4 = 16y^{2} + 24y$. Contradiction! As a result, there are no natural numbers $ a,b,c,d$ satisfies a condition.

Lemmma: the equation $ x^4 - x^2y^2 + y^4 = m^2$ has not solutions in positive integers with x>y. Proof: suppose the opposite, and take x, y $ (x,y) = 1$, with xy minimal, $ (x^2 - y^2)^2 + (xy)^2 = m^2$ $ case1)$if x and y are not both odd, then there exist a and b such that (a,b)=1 and $ xy = 2ab$ and $ x^2 - y^2 = a^2 - b^2$, let $ x = d_1X,b = d_1B,y = d_2Y,a = d_2A,XY = 2AB$ with $ (X,B) = 1 = (Y,A)$, from this there are 2 subcases: $ X = 2A, Y = B$ or $ X = A, Y = 2B$ $ (1)$ $ 4d_1^2A^2 - d_2^2B^2 = d_2^2A^2 - d_1^2B^2$ which is equivalent to $ d_1^2(4A^2 + B^2) = d_2^2(A^2 + B^2)$ crearly $ (A,B) = 1$, let $ d = (4A^2 + B^2,A^2 + B^2)$ d divides $ 3A^2$, but $ A^2 + B^2$ is not congruent to 0 (mod 3), then $ D$ divides $ A^2$, obviously $ D$ divides $ B^2$, but (A,B)=1, then D=1, therefore: $ A^2 + B^2 = C^2$ and $ 4A^2 + B^2 = D^2$, for some C,D. If B were even then settin B=2B_1, we would have two pair of similar equations Then suppose B is odd, then the pitagorean triple gives: $ B = p^2 - q^2$ $ A = pq$, in the first equation: $ p^4 - p^2q^2 + q^4 = C^2$ with $ pq = A < = a < = ab < xy$ a contradiction. $ (2)$ $ d_1^2(A^2 + B^2) = d_2^2(A^2 + 4B^2)$, which is similar $ A = p^2 - q^2$ and $ B = pq$, with $ pq < = b < xy$ $ case2)$ if x and y are both odd, then there exist a and b such that (a,b)=1 and $ xy = a^2 - b^2$ and $ x^2 - y^2 = 2ab$(x>y), from this: $ a^4 - a^2b^2 + b^4 = (\frac {x^2 + y^2}{2})^2$, but these a,b arent both odd, so again in the case 1, there is a contradiction too(or ab=0, in this case x=y, imposibble) Now, for the problem, suppose there is a aritmetic progression: (a,r)=1 $ a(a + r)(a + 2r)(a + 3r) = (a^2 + 3ar)(a^2 + 3ar + 2r^2) = x^2 - (r^2)^2 = n^2$, with $ x = a^2 + 3ar + r^2$, the pitagorean triple again gives 2 cases: $ i)$ $ a^2 + 3ar + r^2 = x^2 + y^2$, $ r^2 = 2xy$, (x,y)=1, this implies $ (a)(a + 3r) = a^2 + 3ar = (x - y)^2$ and $ (a + 2r)(a + r) = a^2 + 3ar + 2r^2 = (x + y)^2$, from (a,r)=1, (a+r,a+2r)=1, and (a+2r,a+3r)= 1 or 3, if is 3, then 3 divides a and doesnt divides r, then (a+r)(a+2r) is congruent to 2 mod3, and is not a perfect square. Contradiction then is 1. From this there are a,b,c,d such that $ a^2 < b^2 < c^2 < d^2$ are an aritmetic progression(wlog not all odd) then $ a^2(2c^2 - b^2) = d^2(2b^2 - c^2)$, and so $ 2(a^2c^2 - b^2d^2) = a^2b^2 - c^2d^2$, setting: $ ac = x, bd = y, ab + cd = 2z, ab - cd = 2w$, then $ x^2 - y^2 = 2zw$, $ xy = z^2 - w^2$, then $ x^4 - x^2y^2 + y^4 = (z^2 + w^2)^2$, but for the lemma there arent solutions or x=y, implying ac=bd>ac, a contradiction. $ ii)$ $ a(a + 3r) = 2y^2$ $ (a + r)(a + 2r) = 2x^2$, (x,y)=1=(a,r), this implies (a,a+3r)=1 or 3, if is 3, then 3 divides a, then (a+r)+(a+2r)=0(mod3), but $ (a + r,a + 2r) = 1$, then $ a + r = g^2$, $ a + 2r = i^2$, obviously: g=i, if is congruent to 0 then 3 divides (x,y) a contradiction; either $ 2x^2 = 1mod 3$, which is impossible. Finally if is 1: a is odd $ a = i^2$ $ a + 3r = 2j^2$, $ 2j^2 - i^2 = 0(mod3)$, then obviously: j=i=0(mod3), then 3 divides a, contradiction because: (a,a+3r)=1. The conclusion follows.

Suppose the progression is x, x+y, x+2y, x+3y. We may assume wlog (x,y)=1. Also: (x,x+y)=1, (x,x+2y)=1or2 and (x,x+3y)=1 or 3 and similarly for the other terms. Therefore either: (x,x+y,x+2y,x+3y)=(a^2,b^2,c^2,d^2) or (a^2,2b^2,c^2,d^2) or (2a^2,b^2,2c^2,d^2) or (3a^2,b^2,c^2,3d^2) or (3a^2,2b^2,c^2,6d^2) or (6a^2,b^2,2c^2,3d^2) where a,b,c,d are pairwise coprime. From Euler, we know the first case has no solutions. By considering the third and fourth modulo 4, and the third modulo 3, we know they have no solutions. The fifth and sixth are, on closer inspection the same, just one of them has negative y. So I'll just look at the fifth. We have c^2=b^2+3d^2 and a^2=b^2-d^2. This latter equation implies either: b+d=k^2, b-d=l^2 OR b+d=2k^2, b-d=2l^2 where k and l are odd. Hence b=1/2[k^2+l^2], d=1/2[k^2-l^2] OR b=k^2+l^2, d=k^2-l^2. In the second case, the first equation gives c^2=4[k^4+l^4-(kl)^2]. This clearly can't work, since it implies c is even which means all the terms of the progression must be even, which contradicts our original assumption. Hence we are left with c^2=(k^2-l^2)^2 + (kl)^2. [k,l are coprime]. This only has solutions if one of k,l is 0 or if they are equal. The first implies that b^2=d^2 hence the solution set is (0,2b^2,4b^2,6b^2) which isn't allowed because the first term is zero. The second implies c^2=a^2 and the same sort of thing applies. I hope that wasn't too brief. Dominic Yeo

Little Gauss wrote: If a,b,c,d satisfies the condition, then so did 2a,2b,2c,2d. Therefore it is enough to consider a case when b-a is even. In this case, we can assume that $ (a,b,c,d) = (x - 3y, x - y, x + y, x + 3y)$ for some natural number $ x,y$. Then $ abcd = (x^{2} - y^{2})(x^{2} - 9y^{2}) = (x^{2} - 5y^{2})^{2} - 16y^{2}$. To be a square number, $ abcd \le (x^{2} - 5y^{2} - 2)^{2}$. ( $ abcd \equiv x^{2} - 5y^{2} \pmod2$) So $ 4(x^{2} - 5y^{2}) - 4 \le 16y^{2}$ But we know that $ x - 3y = a$ greater than 0. So we get $ 16y^{2} \ge 4((3y + 1)^{2} - 5y^{2}) - 4 = 4(4y^{2} + 6y + 1) - 4 = 16y^{2} + 24y$. Contradiction! As a result, there are no natural numbers $ a,b,c,d$ satisfies a condition. Your solution is wrong. $(x^{2} - y^{2})(x^{2} - 9y^{2}) = (x^{2} - 5y^{2})^{2} - 16y^{4}$ And not $-16y^{2}$