Answer to Question #350946 in Combinatorics | Number Theory for WSN

Question #350946

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


1
Expert's answer
2022-06-28T15:15:37-0400

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.


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