Question

3. Prove that there exists infinitely many positive integers n such that 4n²+ 1 is divisible both by 5 and 13. 

Answer 1

Divisibility both by 5 and 13 means divisibility by 65. Consider infinitely many positive integers of the form $n = 65k + 4, \ k \in \mathbb{N}$. Then

$4n^2 + 1 = 4(65k + 4)^2 + 1 = 4(4225k^2 + 520k + 16) + 1 = 16900k^2 + 2080k + 65 = 65(260k^2 + 32k + 1)$is divisible by 65, hence it is divisible both by 5 and 13.