Solve the following equation in natural numbers: x2=2y+2021z
Problem
Source: Serbia JBMO TST 2021
Tags: number theory, modular arithmetic, Diophantine Equations, collection
03.01.2022 23:25
I claim (x,y,z)=(45,2,1) is the only solution. Note that if y=1 then x^2\equiv 3\pmod{4}; thus y\ge 2. Assume y=2. We have (x-2)(x+2) = 43^z \cdot 47^z. Noticing (x-2,x+2)=1, it follows x-2=43^z and x+2=47^z. That is, 47^z - 43^z=4. From here, we immediately obtain z=1, yielding the triple above. Now assume y\ge 3. Using modulo 8, it follows z is even, set z=2k with k\ge 1 to arrive at \bigl(x-2021^k\bigr)\bigl(x+2021^k\bigr) = 2^y. In particular, (a) x\pm 2021^k are both even; and (b) if d\mid x\pm 2021^k then d\mid 2\cdot 2021^k. Consequently, d=2. That is, we must have 2\cdot 2021^k + 2 = 2^{y-1} \implies 2021^k + 1 = 2^{y-2}. Clearly, y>4. Inspecting modulo 4, however, this is a clear contradiction.
22.01.2022 01:38
grupyorum wrote: I claim (x,y,z)=(45,2,1) is the only solution. Note that if y=1 then x^2\equiv 3\pmod{4}; thus y\ge 2. Assume y=2. We have (x-2)(x+2) = 43^z \cdot 47^z. Noticing (x-2,x+2)=1, it follows x-2=43^z and x+2=47^z. That is, 47^z - 43^z=4. From here, we immediately obtain z=1, yielding the triple above. Now assume y\ge 3. Using modulo 8, it follows z is even, set z=2k with k\ge 1 to arrive at \bigl(x-2021^k\bigr)\bigl(x+2021^k\bigr) = 2^y. In particular, (a) x\pm 2021^k are both even; and (b) if d\mid x\pm 2021^k then d\mid 2\cdot 2021^k. Consequently, d=2. That is, we must have 2\cdot 2021^k + 2 = 2^{y-1} \implies 2021^k + 1 = 2^{y-2}. Clearly, y>4. Inspecting modulo 4, however, this is a clear contradiction. why is z an even number?
22.01.2022 09:43
Because if z is odd number then x^2\equiv 0+5^z\equiv 5^{2k+1}\equiv(25)^k\cdot 5\equiv 5\pmod{8} which is absurd.
10.08.2022 19:10
What is d?
07.03.2024 10:12
d is a random divisor of x+-2021^k and 2*2021^k