1. The problem states that the $r$th term of the first sequence is $2r^2 - 1$ and the $r$th term of the second sequence is $40 - r^2$. We need to show there is only one number common to both sequences. 2. To find common terms, set the $r$th terms equal: $$2r^2 - 1 = 40 - r^2$$ 3. Rearrange the equation to isolate terms on one side: $$2r^2 - 1 - 40 + r^2 = 0$$ $$3r^2 - 41 = 0$$ 4. Solve for $r^2$: $$3r^2 = 41$$ $$r^2 = \frac{41}{3}$$ 5. Since $r$ must be an integer (as it represents the term position in the sequence), check if $\frac{41}{3}$ is a perfect square integer. It is not. 6. This means no integer $r$ satisfies the equality for the same $r$ in both sequences. 7. However, the problem implies the sequences might share a common number at different positions. Let the first sequence term be $2m^2 - 1$ and the second be $40 - n^2$ for integers $m,n$. 8. Set: $$2m^2 - 1 = 40 - n^2$$ $$2m^2 + n^2 = 41$$ 9. We look for integer solutions $(m,n)$ to this Diophantine equation. 10. Test integer values of $n$ from 0 to 6 (since $n^2$ must be less than or equal to 41): - For $n=5$, $n^2=25$, then $2m^2 = 41 - 25 = 16$, so $m^2 = 8$ (not a perfect square). - For $n=3$, $n^2=9$, then $2m^2 = 41 - 9 = 32$, so $m^2 = 16$, which is a perfect square ($m=4$). 11. So $m=4$, $n=3$ is a solution: $$2(4)^2 - 1 = 2 \times 16 - 1 = 32 - 1 = 31$$ $$40 - (3)^2 = 40 - 9 = 31$$ 12. The common number is 31. 13. Check other $n$ values to confirm no other solutions exist. 14. Therefore, there is only one number, 31, common to both sequences. Final answer: The only common number in both sequences is $31$.