if p is prime then p^2 = 1 mod 24

Problem:
Prove that for $p > 3$ prime, $p^2 \equiv 1 \pmod{24}$ .

Solution:
Consider $p^2 - 1 = (p-1) (p+1)$ .
Since $p > 3$ , we know that either $p-1$ or $p+1$ is divisible by 3. Both are divisible by 2. Also, one of them is divisible by 4, since p is either $-1$ or $1 \pmod 4$ . Hence $p^2-1$ is divisible by $2 \times 3 \times 4 = 24$ .

QED

Comments

p - p^2-2 / p+2

J595 Mathematical Reflection

$\sqrt[3]{(x-1)^2} - \sqrt[3]{2(x-5)^2} + \sqrt[3]{(x-7)^2} = \sqrt[3]{4x}$ By Titu Adreescu. My Solution (potentially wrong): Let

$A = \sqrt[3]{(x-1)^2}$ ,

$B = \sqrt[3]{2(x-5)^2}$ ,

$C = \sqrt[3]{(x-7)^2}$ ,

$D = \sqrt[3]{4x}$ . Hence we have

$A + C = B + D$ . Notice that

$A^3 + C^3 = B^3 + D^3$ . Also notice that

$(A+C)^3 = (B+D)^3$ yields

$AC = BD$ . Also by factoring out

$A^3 + C^3 = (A+C)(A^2-AC-C^2)$ and doing the same for

$B^3+D^3$ gives us

$A^2 + C^2 = B^2 + D^2$ . Lemma 1 :

$(A^3 - C^3)^2 = (B^3 - D^3)^2$ . Proof: Note that

$(A^3+C^3)^2 - (A^3-C^3)^2 = 4A^3C^3 = 4B^3D^3 = (B^3+D^3)^2 - (B^3-D^3)^2$ and the conclusion follows.

$\square$ Using Lemma 1 we get

$( (x-1)^2 - (x-7)^2)^2 = (2(x-5)^2 - 4x)^2$ and solving for x gives us

$x = (9 \pm 4\sqrt{2}) , (3 \pm 2\sqrt{2})$ .

$\square$

232 | 20^n + 16^n - 3^n - 1

Problem: If

$n$ is even, then

$323 | 20^n + 16^n - 3^n - 1$ . Solution: Let

$n = 2k$ . We have

$17 | 400^k - 9^k$ and

$17 | 256^k - 1$ . Hence

$17 | 20^n + 16^n - 3^n - 1$ . Also

$19 | 20^n -1$ and

$19 | 256^k - 9^k$ . Hence

$19 | 20^n + 16^n - 3^n - 1$ . Hence it is divisible by

$17 \times 19 = 323$ . QED

Maths by Chance

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