nicetry007 wrote: Let $ 2d-1 = x^{2}$, $ 5d-1 = y^{2}$ and $ 13d-1 = z^{2}$. Suppose $ d$ is even. This would imply $ 2d-1 = 3 ( mod \; 4)$. But a perfect square is either $ 0$ or $ 1 (mod \; 4)$. Hence, $ d$ is odd. This implies that $ x$ is odd and $ y$ and $ z$ are even. $ y^{2}-z^{2}= 8d \Rightarrow y^{2}= z^{2}(mod \;8)$. Since the perfect square of an even number is either $ 0$ or $ 4 (mod \; 8)$, $ y^{2}$ and $ z^{2}$ are either both $ 0(mod\; 8)$ or $ 4 (mod \; 8)$. Suppose $ y^{2}= 0 (mod \; 8)$ and $ z^{2}= 0 (mod \; 8)$. If the square of an integer is divisible by $ 8$, then it is also divisible by $ 16$. Hence, it follows that $ y^{2}= 0 (mod \; 16)$ and $ z^{2}= 0 (mod \; 16)$. This, in turn, implies that $ y^{2}-z^{2}= 0 (mod \;16)$. But $ y^{2}-z^{2}= 8d$ and $ d$ is odd. Hence, we arrive at a contradiction. Suppose $ y^{2}= 4 (mod \; 8)$ and $ z^{2}= 4 (mod \; 8)$. Let $ y^{2}= 4p^{2}$ and $ z^{2}= 4q^{2}$ where $ p,q$ are odd. $ y^{2}-z^{2}= 4(p^{2}-q^{2}) = 4(p-q)\;(p+q)$. Since $ (p-q)$ and $ (p+q)$ are both even, $ y^{2}-z^{2}$ is divisible by $ 16$. But $ y^{2}-z^{2}= 8d$ and $ d$ is odd. We observe that in both cases we arrive at a contradiction. Hence, the result follows.

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We see that $y$ and $z$ have the same parity since $8$ divides $y^2-z^2$.But we see that $y$ must be of the form $5k+/-2$ and $z$ must be of the form $13k+/-5$.Here we see that $y$ and $z$ have different parity.Hence

This is a problem of IMO(1986)

let assume $2d-1=x^2,5d-1=y^2,13d-1=z^2 \implies$ $x^2$ is odd so $x^2=2d-1 \equiv 1 \mod 8 \implies d \equiv 1 \mod 4$ now let $d=4k+1$ which implies that $8k+1=x^2,20k+4=y^2,52k+12=z^2$ now let: $z=2z_0,y=2y_0 \implies 5k+1={y_0}^2,13k+3={z_0}^2$, we shall assert $5k+1,13k+3$ cannot be perfect square at the same time and for proving this we check four cases: case 1: $k \equiv 0 \mod 4 \implies {z_0}^2 \equiv 3 \mod 4$ which is impossible case 2: $k \equiv 1 \mod 4 \implies {y_0}^2 \equiv 2 \mod 4$ which is impossible case 2: $k \equiv 2 \mod 4 \implies {y_0}^2 \equiv 3 \mod 4$ which is impossible case 2: $k \equiv 3 \mod 4 \implies {z_0}^2 \equiv 2 \mod 4$ which is impossible