Prove that there exists infinitely many positive integers n such that 4n^2 + 1 is divisible both by 5 and 13.	Select:	A) 5	B) 6	C) 5|4n^2+ 1 and 13|4n^2+ 1	D) 3|4n^2+ 1 and 10|4n^2+ 1	E) 2|4n^2+ 1 and 1|4n^2+ 1
For instance, all numbers n in the arithmetic progression 65k+56 (k = 0, 1, 2, ... ) have the desired property. Indeed, if n = 65k+56 with an integer k&gt;=0, then n == 1 (mod 5) and n == 4 (mod 13), hence 4n^2+ 1 == o (mod 5) and 4n^2+ 1== 0 (mod 13). Thus, 5|4n^2+ 1 and 13|4n^2+ 1   Answer is 5|4n^2+ 1 and 13|4n^2+ 1 5|4n^2+ 1 and 13|4n^2+ 1
<result>C</result>
Prove that there exists infinitely many positive integers n such that 4n^2 + 1 is divisible both by 5 and 13.