see $(mod.16)$ the numbers. the quadratic residues $(mod.16)$ are $0$,$1$,$4$,$9$. suppose that $2d-1$,$5d-1$,$13d-1$ are perfect squares. for $2d-1\equiv 0,1,4,9 (mod.16)$, we have $d \equiv 1,5,8 or 13 (mod.16)$. if $d$ is $1$ or $13$, then $13d - 1$ is not one of these values. If $d$ is $5$ or $9$, then $5d - 1$ is not one of these values. contradiction.

Assume $ 2d,5d,13d$ are all perfect squares. So $ d$ is divisible by $ 2,5,13$. Let $ d=130n$. Note that $ d/2,d/5,d/13$ are perfect squares, let the values be $ p^2,q^2,r^2$ respectively. So $ 65n=p^2,26n=q^2,10n=r^2$. Multiply the three equations, we get $ 130^2n^3=(pqr)^2$. Therefore $ n$ is a perfect square. But this implies that $ 2d=260n$ is not a perfect square. Contradiction.

Johan Gunardi wrote: Assume $ 2d,5d,13d$ are all perfect squares. So $ d$ is divisible by $ 2,5,13$. Let $ d = 130n$. Note that $ d/2,d/5,d/13$ are perfect squares, let the values be $ p^2,q^2,r^2$ respectively. So $ 65n = p^2,26n = q^2,10n = r^2$. Multiply the three equations, we get $ 130^2n^3 = (pqr)^2$. Therefore $ n$ is a perfect square. But this implies that $ 2d = 260n$ is not a perfect square. Contradiction. We have to assume $ 2d-1, 5d-1, 13d-1$ are all perfect squares.

whoops.. sorry, i was uncareful.

it is IMO1986 first prob..and it is also exist in pen http://www.mathlinks.ro/viewtopic.php?t=150420

Assume $ 2d-1, 5d-1, 13d-1$ are perfect squares Then $ (2d-1)(5d-1)(13d-1) = x^2$ for some $ x \ge 0$ $ (2d-1)(5d-1)(13d-1) = 130 d^3-101 d^2+20 d-1$ $ 130 d^3-101 d^2+20 d-1 \equiv -d^2-1 \bmod 5$ The residues $ \bmod$5 are $ {1,4}$ Then $ -d^2-1 \equiv -2, 0 \bmod 5$, but since this equals $ x^2$ the residues$ \bmod 5$, have to match. Contradiction.

Complex_Ninja wrote: Assume $ 2d-1, 5d-1, 13d-1$ are perfect squares Then $ (2d-1)(5d-1)(13d-1) = x^2$ for some $ x \ge 0$ $ (2d-1)(5d-1)(13d-1) = 130 d^3-101 d^2+20 d-1$ $ 130 d^3-101 d^2+20 d-1 \equiv -d^2-1 \bmod 5$ The residues $ \bmod$5 are $ {1,4}$ Then $ -d^2-1 \equiv -2, 0 \bmod 5$, but since this equals $ x^2$ the residues$ \bmod 5$, have to match. Contradiction. I dont see the contradiction in your argument.

-2 mod 5 is not equal to 1 or 4 mod 5 and 0 mod 5 is not equal to 1 or 4 mod 5. since x^2 is a perfect square it is either 1 or 4 mod 5. but in this case it is -2 which is also 3 or 0 mod 5. so contradiction. edit: oops. dont know how i missed that 0 is not a quadratic residue mod 5. sorry.

And if $x^2$ is divisible by 5?

I don't mean to revive this topic, but I was wondering if this was a valid proof (I'm pretty sure it is, but just in case I missed something...):

Perfectly correct, flare. If you will take a look at the link mentioned above your post (pointing to the problem in PEN), you will see a solution very much along your line.

To the contrary let us assume that $13d-1,5d-1,2d-1$ are squares and let $13d-1=k^2$....(1) $5d-1=l^2$.......(2) $2d-1=m^2$.....(3) (1) $\Rightarrow 13d \equiv 1,2\pmod{4} \Rightarrow d \equiv 1,2\pmod{4}$. $d \equiv 1\pmod{4} \Rightarrow d=4k+1$ for some $k \in \mathbb{Z}$.Then (2) $\Rightarrow 20k+4=4(5k+2)=l^2 \Rightarrow (\frac{l}{2})^2 \equiv 3\pmod{5}$.Contradiction! $d \equiv 2\pmod{4} \Rightarrow d=4k+2$ for some $k \in \mathbb{Z}$.Then (3) $\Rightarrow m^2 \equiv 3\pmod{8}$.Contradiction! We have acheived a contradiction!

Taking $A=a^2, B=b^2, C=c^2$ in your system, it does not easily solve to give $A=B=C=-1$. The system is indeterminate (compute its determinant, and see it is null). The general solution is $C=X, B = \dfrac {5C - 8} {13}, A = \dfrac {10C - 55} {65}$, and indeed it has the particular solution $A=B=C=-1$ for $X=-1$, but it also has infinitely many others (in rationals), and you have to show that none yields $A,B,C$ perfect squares.

Is this valid:

Assume that there is exist $d$ such that for every 2 distinct $a,b \in \{2, 5, 13, d\}$, $ab-1$ is perfect square. So there are exist $a,b,c\in\mathbb{N}$ such that $2d-1=a^2$, $5d-1=b^2$, $13d-1=c^2$. Clearly see that $a$ is odd so $2d-1\equiv 1\pmod 4$ that means $d$ is odd too. So, $b,c$ is even. There are exist $r,s\in\mathbb{N}$ such that $2^r\parallel b$ and $2^s\parallel c$ If $r<s$ then $2^{2r}\parallel b^2-c^2$. Because $2^3\parallel 8d$ and $b^2-c^2=8d$. So $2r=3$ that is impossible. If $r>s$ then $2^{2s}\parallel b^2-c^2$. In the same way, we get $2s=3$ that is impossible too. If $r=s$ then assume that $b=2^rm,c=2^rn$ when $m,n$ is odd. So $8d=b^2-c^2=2^{2r}(m^2-n^2)$ Because $m^2\equiv n^2\equiv 1 \pmod 8$. So $8\mid m^2-n^2$. That means $2^{2r}$ is odd so $r=0$ that is impossible. Contradiction.

supercomputer wrote: Is this valid:

I don't think this works because the first term can be $1\pmod{4}$ and the other two can be $0\pmod{4}$ in the set $\{8k+1,4(5k+1),4(13k+3)\}$.

Providing with a very beginner solution (which I solved in my Art Period) : SOLUTION

This claim is only used in the whole solution. We note down all possiblities of $ab-1$ where $a,b \in (2,5,13,d)$: $(9,64,25)$ and $(2d-1,5d-1,13d-1)$ The first set is a perfect square. Let us have a look at the second set. Assuming all the elements of the second set are perfect squares, we get: $2d-1=a^2$ $(\clubsuit)$ $5d-1=b^2$ $(\bigstar)$ $13d-1=c^2$ $(\spadesuit)$ We know the LHS of $(\clubsuit)$ is an odd number. So, $2d-1=8k+1\Rightarrow d=4k+1$ Replacing values of $d$ in $(\bigstar)$ and $(\spadesuit)$, we get: $20k+4=4(5k+1)$ and $52k+12=4(13k+3)$. Thus, $(5k+1) \equiv (k+1) \equiv {0,1}\pmod{4}$. so, from $(\bigstar)$, we get $k \equiv {-1,0}\pmod4$. Similarly, $(13k+3) \equiv (k-1) \equiv {0,1}\pmod{4}$. so, from $(\bigstar)$, we get $k \equiv {1,2}\pmod4$. But, these two cannot hold together for any $d\in\mathbb{N}$. So, our assumption is wrong. $\boxed{Q.E.D}$

This problem has been the bane of my existence for far too long. I guess it's better late than never.... It suffices to show that at least one of $2d-1$, $5d-1$, or $13d-1$ is not a perfect square. Suppose for the sake of contradiction that all of them are. First observe that since $2d-1$ is a perfect square, $d$ must be odd (else $2d-1\equiv 3\pmod 4$). Furthermore, $d$ divides each of $a^2 + 1$, $b^2 + 1$, and $c^2 + 1$; thus $d\equiv 3\pmod 4$ is impossible as well. It follows that $d\equiv 1\pmod 4$. So let $d = 4k + 1$. Substitution implies that \[ \frac{5d-1}4 = \frac{5(4k+1) - 1}4 = 5k + 1\quad\text{and}\quad \frac{13d-1}4 = \frac{13(4k+1) - 1}4 = 13k + 3 \]are both perfect squares. But now observe that these two numbers differ by $8k + 2$, which is an integer that is $2\pmod 4$; thus these two numbers cannot simultaneously be perfect squares, and we are done.

Nice one....

QED

Does mod 7 work?

We will show that amongst $2d-1,5d-1,13d-1$ there is a number that is not a perfect square. For the sake of contradiction assume that there doesn't, and let \[p^2 = 2d-1;\;q^2 =5d-1;\;r^2 = 13d-1\]we then have that \[\frac{p^2+1}{2} = \frac{q^2+1}{5}\iff 5p^2+5 = 2q^2+2\iff 2q^2-5p^2=3\]However, running through the fact that $x^2\in\{0,1\}\mod{4}$ we have that this is impossible. Thus we have a contradiction and we are done.

DrYouKnowWho wrote: we then have that \[\frac{p^2+1}{2} = \frac{q^2+1}{5}\iff 5p^2+5 = 2q^2+2\iff 2q^2-5p^2=3\]However, running through the fact that $x^2\in\{0,1\}\mod{4}$ we have that this is impossible. Thus we have a contradiction and we are done. Something must be all wrong here, since $(p,q)=(1,2)$ actually is a solution.

I did with mod $3$. I am not sure of it though

Let's assume that ab-1 is a square We know that all perfect squares are either of form 4k or 8k+1 (i) If ab-1 =4k then ab=4k+1 But this isn't possible since for a=2, ab is even (ii)If ab-1=8k+1 then ab=8k+2 This implies d is odd as 2d=2(4k+1) But then 5d ,13d are odd, making ab is odd Hence we have a contradiction in each case, our assumption is wrong Therefore we can always have d in the set for which ab-1 is not a perfect square. Was just wondering if this solution works... can anyone figure out any faults or shortcomings?

Suppose not. Then $2d-1$, $5d-1$, and $13d-1$ are all perfect squares. We note that the quadratic residues $\pmod {16}$ are $0,1,4,9$. Case 1: $d$ is even. Then $2d-1\equiv 3\pmod4$, and thus not a perfect square. Case 2: $d$ is odd. Then $5d-1$ and $13d-1$ are both even perfect squares. Since they are distinct $\pmod {16}$, either $8d\equiv 4\pmod {16}$ or $8d\equiv 12\pmod{16}$. Both are not possible as they both imply $8\nmid 8d$.

Suppose FTSOC that $2d-1,5d-1,$ and $13d-1$ are all perfect squares; therefore, they are quadratic residues modulo $4.$ Then, $2d-1\equiv 0,1\pmod{4}$ or $d\equiv 1,3\pmod{4}.$ Also, $5d-1\equiv d-1\equiv 0,1\pmod{4}$ or $d\equiv 1,2\pmod{4}.$ Hence, $d\equiv 1\pmod{4},$ so let $d=4d'+1.$ We can then let $b^2=5d-1=20d'+4$ and $c^2=13d-1=52d'+12.$ Since $2\mid b,c,$ let $2b_1=b$ and $2c_1=c,$ and we know that $b_1^2=5d'+1$ and $c_1^2=13d'+3.$ Subtracting yields $$c_1^2-b_1^2\equiv 8d'+2\equiv 2\pmod{8}$$which is not a difference of quadratic residues. $\square$

Suppose $2d-1, 5d-1, 13d-1,$ are squares. Then they are congruent to $0$ or $1$ in modulo $4.$ Notice that $2d-1$ implies $d$ is odd. Setting $d=2k+1\implies 10k+4=d^2\implies b,k$ is even. Setting $k=2h \implies 5k+1$ is a square. Similarly we get that $13h+3$ is a square. We can easily get a contradiction from the difference of two squares, $$(13h+3)-(5h+1)=8h+2. \text{Q.E.D.}$$

Here is my solution: https://calimath.org/pdf/IMO1986-1.pdf And I uploaded the solution with motivation to: https://youtu.be/JS6xuvdYd5Y

here we go: FTSOC assume $\exists$ $a,b \in \{2,5,13,d\}$ for which $ab-1$ is a perfect square, then: $\alpha^2=2d-1$ $\beta^2=5d-1$ $\gamma^2=13d-1$ now $\alpha^2 \equiv 1 \pmod 4 \implies 2d-1=4m+1 \implies d=2m+1$ which is odd now plugging this $d$ back into the $\beta^2$ and $\gamma^2$ we get $\beta^2=10m+4$ and $\gamma^2=26m+12$ now clearly this gives us that $\beta$ and $\gamma$ are even so set $\beta=2x$ and $\gamma=2y$ so we get $4(x^2-y^2)=8d \implies x^2-y^2 \equiv 2 \pmod 4$ which is a contradiction! as there is no $(x,y) \in \mathbb{Z}$ s.t. $x^2-y^2 \equiv 2 \pmod 4$ and hence we get $\exists$ $a,b \in \{2,5,13,d\}$ such that $ab-1$ is not a perfect square $\blacksquare$

Suppose FTSOC that there exists $d$ such any pair produces a perfect square. Then all three of $2d-1$, $5d-1$, $13d-1$ are perfects squares. We must have $2d-1\equiv 1 \pmod 4\Rightarrow d$ is odd, then $5d-1$ is even, and so $5d-1\equiv 0 \pmod 4 \Rightarrow d\equiv 1 \pmod 4$. Let $d=4k+1$ for some integer $k$. Inserting it into $5d-1$ and $13d-1$, we get that $20k+4$ and $52k+12$ are perfect squares, respectively, and they should remain such after being divided by $4$, i.e. $5k+1$ and $13k+3$ are perfect squares, but $$5k+1\equiv 0,1 \pmod 4\Rightarrow k\equiv 0,3 \pmod 4$$$$13k+3\equiv 0,1 \pmod 4\Rightarrow k\equiv 1,2\pmod 4$$which is a desired contradiction.

Assume for the sake of contradiction that there exist $k_1^2=2d-1$, $k_2^2=5d-1$, $k_3^2=13d-1$ for positive integers $k_1$, $k_2$, $k_3$. Since squares are only 0, 1, 4 mod 8, we must have $2d-1 \equiv 1 \pmod 8 \implies d \equiv 1 \pmod 4$. Let $d=4a+1$. Substituting, we get $k_1'^2=8a+1$, $k_2'^2=5a+1$, $k_3'^2=13a+3$ for some positive integers $k_1'$, $k_2'$, $k_3'$. Then, note that $k_3'^2-k_2'^2=8k+2 \equiv 2 \pmod 4$, but this is impossible since squares are only 0, 1 modulo 4, contradiction. $\blacksquare$

I did differently from of all you guys, let me show: FTSOC, suppose that all the combinations we try give us a perfect square. Firstly, if none of $a$ or $b$ is $d$, obviously give us all perfect squares: $ab-1=\begin{cases} 2\cdot 5-1= 3^2 \\2\cdot13-1= 5^2 \\ 5\cdot 13-1= 8^2\end{cases}$ are all perfect squares. Now, assume WLOG, $b=d$. Then: $ab-1= \begin{cases} 2d-1= q_1^2&\text{(1)}\\5d-1=q_2^2&\text{(2)}\\13d-1=q_3^2&\text{(3)} \end{cases}$, where $q_1, q_2,q_3\in\mathbb{Z^*_+}$ Now, multiply (1) by $65$, (2) by $26$ and (3) by $10$: $\begin{cases} 130d-65= 65q_1^2\\130d-26=26q_2^2\\130d-10=10q_3^2 \end{cases}$ So, $130d=65(q_1^2+1)= 26(q_2^2+1)= 10(q_3^2+1)$. In particular, $26(q_2^2+1)= 10(q_3^2+1)\implies 13(q_2^2+1)= 5(q_3^2+1)$ As $13\nmid 5\implies q_3^2+1= 13\implies q_3^2= 12\implies q_3\not\in\mathbb{Z}$, contradiction! Therefore, our supposition is false and we can always choose $a,b\in\{2,5,13,d\}$ such $(ab-1)$ is not a perfect square.

Assume for the sake of contradiction that the statement is false. Then, we have\begin{align*} 2d-1&=c^2 \\ 5d-1&=f^2 \\ 13d-1&=g^2 \\ \end{align*}for some integers $c,$ $f,$ and $g.$ Note that $65c^2+65=130d=26f^2+26=10g^2+10.$ Since $c^2\equiv 1 \pmod{2},$ $c\equiv 1 \pmod{2},$ so $c^2\equiv 1\pmod{4}.$ As such, $10g^2+10\equiv 26f^2+26\equiv 130\equiv 2\pmod{4},$ so $5g^2+5\equiv 13f^2+13\equiv 1 \mod{2},$ so $g\equiv f\equiv 0\pmod{2}.$ Denote $f=2h$ and $g=2k$ for some integers $h,k.$ Note that $5+5c^2=2+2f^2,$ and $13+13c^2=2+2g^2,$ so $8c^2+8=2g^2-2f^2,$ so $4c^2+4=g^2-f^2=4k^2-4h^2,$ and $k^2-h^2=c^2+1\equiv 2 \pmod{4},$ a contradiction, so the statement is true. $\Box$