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By J J Sakurai

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Thus the last digit is 3. 184 Example Prove that every year, including any leap year, has at least one Friday 13th. Solution: It is enough to prove that each year has a Sunday the 1st. ) Now, each remainder class modulo 7 is represented in the third column, thus each year, whether leap or not, has at least one Sunday the 1st. 185 Example Find infinitely many integers n such that 2n + 27 is divisible by 7. Solution: Observe that 21 ≡ 2, 22 ≡ 4, 23 ≡ 1, 24 ≡ 2, 25 ≡ 4, 26 ≡ 1 mod 7 and so 23k ≡ 1 mod 3 for all positive integers k.

Solution: There are no such solutions. All perfect fourth powers mod 16 are ≡ 0 or 1 mod 16. This means that n41 + · · · + n414 can be at most 14 mod 16. But 1599 ≡ 15 mod 16. 190 Example (P UTNAM 1986) What is the units digit of 1020000 ? 10100 + 3 52 Chapter 3 Solution: Set a − 3 = 10100. Then [(1020000)/10100 + 3] = [(a − 3)200/a] = 1 200 200 200−k 200 k 200 199−k = (−3)k. Since 200 [ a (−3)k] = 199 k=0(−1) k=0 k a k a k=0 k 200 199−k k 200 (−3)k ≡ = −3199. As a ≡ 3 mod 10, 199 0, (3)199 199 k=0 k a k=0(−1) k k 200 ≡ −3199 ≡ 3 mod 10.

Each factor is greater than 1 for n > 1, and so n4 + 4 cannot be a prime. 39 Some Algebraic Identities 133 Example Find all integers n ≥ 1 for which n4 + 4n is a prime. Solution: The expression is only prime for n = 1. Clearly one must take n odd. For n ≥ 3 odd all the numbers below are integers: n4 + 22n = n4 + 2n22n + 22n − 2n22n 2 = (n2 + 2n)2 − n2(n+1)/2 = (n2 + 2n + n2(n+1)/2)(n2 + 2n − n2(n+1)/2). It is easy to see that if n ≥ 3, each factor is greater than 1, so this number cannot be a prime.

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