Prime Power Divisibility in a Sum of Squares
A number theory problem asks you to determine the largest power of 3 dividing N = 5^{2024} + 7^{2024}. A classmate claims that since 5 + 7 = 12 is divisible by 3, and the exponents are even, N must be divisible by 3^2 but not by 3^3. You decide to check this using modular arithmetic and the binomial theorem. You recall that for odd primes p, the exponent of p in a sum of like powers can sometimes be found by factoring. Here the exponents are even, so you consider writing N as a difference of squares or using the fact that 5 ≡ -1 mod 3 and 7 ≡ 1 mod 3. You also note that 2024 is even, so 5^{2024} ≡ 1 mod 3 and 7^{2024} ≡ 1 mod 3, giving N ≡ 2 mod 3, which would contradict the classmate's claim. You need to resolve this discrepancy and find the exact 3-adic valuation of N.
1. Compute N modulo 3 and modulo 9 to test the classmate's claim. What does this reveal about the divisibility of N by 3?
Read the worked answer
Since 5 ≡ 2 mod 3 and 7 ≡ 1 mod 3, we have 5^{2024} ≡ 2^{2024} mod 3. Because 2024 is even, 2^{2024} ≡ (2^2)^{1012} ≡ 1^{1012} ≡ 1 mod 3. Also 7^{2024} ≡ 1^{2024} ≡ 1 mod 3. Thus N ≡ 1 + 1 = 2 mod 3. Therefore N is not divisible by 3 at all. The classmate's claim that N is divisible by 9 is false. Checking modulo 9: 5 ≡ 5, 5^2 = 25 ≡ 7, 5^3 ≡ 5·7=35 ≡ 8, 5^4 ≡ 5·8=40 ≡ 4, 5^5 ≡ 5·4=20 ≡ 2, 5^6 ≡ 5·2=10 ≡ 1 mod 9. So the order of 5 mod 9 is 6. Since 2024 = 6·337 + 2, 5^{2024} ≡ 5^2 ≡ 7 mod 9. Similarly 7 ≡ 7, 7^2=49 ≡ 4, 7^3 ≡ 7·4=28 ≡ 1 mod 9, so order 3. 2024 = 3·674 + 2, so 7^{2024} ≡ 7^2 ≡ 4 mod 9. Then N ≡ 7+4=11 ≡ 2 mod 9. So N ≡ 2 mod 9, confirming N is not divisible by 3. The classmate confused the sum of bases with the sum of powers.
2. Find the exact 3-adic valuation v_3(N). Justify your answer using modular arithmetic or factoring.
Read the worked answer
From the previous computation, N ≡ 2 mod 3, so v_3(N) = 0. The number N is not divisible by 3. To see this structurally, note that modulo 3, 5 ≡ -1 and 7 ≡ 1, so 5^{2024} + 7^{2024} ≡ (-1)^{2024} + 1^{2024} = 1 + 1 = 2 mod 3. Since 2 is not 0 mod 3, 3 does not divide N. Therefore the largest power of 3 dividing N is 3^0 = 1. The classmate's reasoning failed because they incorrectly assumed that a ≡ b mod 3 implies a^k ≡ b^k mod 3 in a way that preserves the sum of bases, but actually the sum of the bases being divisible by 3 does not imply the sum of the powers is divisible by 3. In fact, for even exponent, the sum of two numbers congruent to 1 and -1 mod 3 is 2 mod 3.