Number Theory Proof of the day--11/23

[Steven G]

Show that x^2+1=0 (mod p), where p is prime, has a solution iff p=1 (mod 4)

 

[Harry_the_cat]

Show that x^2+1=0 (mod p), where p is prime, has a solution iff p=1 (mod 4)

Is this question complete or is something missing?

Suppose \(\displaystyle p=7=0(mod7)\), ie \(\displaystyle p\) is prime.

\(\displaystyle x^2+1 = 7\) has a solution BUT \(\displaystyle p\neq1(mod 4)\).

Contradiction! Disproven.

 

[BigBeachBanana]

\(\displaystyle x^2+1 = 7\) has a solution BUT \(\displaystyle p\neq1(mod 4)\).

[imath]x^2+1 \equiv 0 \mod 7[/imath] has no integer solutions, but [imath]p \equiv 5 \equiv 1 \mod 4[/imath] does.

A counter example would be [imath]p=2[/imath], but the question is valid for odd primes.

 

[Steven G]

Is this question complete or is something missing?

Suppose \(\displaystyle p=7=0(mod7)\), ie \(\displaystyle p\) is prime.

\(\displaystyle x^2+1 = 7\) has a solution BUT \(\displaystyle p\neq1(mod 4)\).

Contradiction! Disproven.

Can you please state your solution?
Actually just consider the following:
0^2 + 1 mod 7 = 1 mod 7
1^2 + 1 mod 7 = 2 mod7
2^2 + 1 mod 7 = 5 mod7
3^2 + 1 mod 7 = 3 mod7
4^2 + 1 mod 7 = 3 mod7
5^2 + 1 mod 7 = 5 mod7
6^2 +1 mod 7 = 2 mod7
Therefore x^2 + 1 = 0 mod 7 has no solution.

 

[Harry_the_cat]

[imath]x^2+1 \equiv 0 \mod 7[/imath] has no integer solutions, but [imath]p \equiv 5 \equiv 1 \mod 4[/imath] does.

A counter example would be [imath]p=2[/imath], but the question is valid for odd primes.

Who said anything about integer solutions? That's why I asked if the question is complete.

Then shouldn't the question read "Show that x^2+1=0 (mod p), where p is an odd prime, has an integer solution iff p=1 (mod 4)."

 

[BigBeachBanana]

Who said anything about integer solutions? That's why I asked if the question is complete.

Then shouldn't the question read "Show that x^2+1=0 (mod p), where p is an odd prime, has an integer solution iff p=1 (mod 4)."

Number theory is the study of integers, thus it implicitly implies integer solutions. However, I agree it should explicitly read "odd prime".

 

[Harry_the_cat]

Number theory is the study of integers, thus it implicitly implies integer solutions. However, I agree it should explicitly read "odd prime".

Yeah fair enough.

 

[Steven G]

Who said anything about integer solutions? That's why I asked if the question is complete.

Then shouldn't the question read "Show that x^2+1=0 (mod p), where p is an odd prime, has an integer solution iff p=1 (mod 4)."

Modulo arithmetic is all about integers.

 

[blamocur]

Modulo arithmetic is all about integers.

[imath]\pi \equiv 0.14159265358979312... \;mod\; 1[/imath]

 

[khansaheb]

Modulo arithmetic is all about integers

Has that been stated in any textbook/reference - explicitly.

We can do modulo type of calculation with area of a sphere (involving π) and area of a circle (involving π} and get an integer output.

Thus can we say [(surface area of a sphere of radius 2)|(surface area of a circle of radius 2)]

 

[blamocur]

[imath]\pi \equiv 0.14159265358979312... \;mod\; 1[/imath]

...i.e. [imath]x \equiv y \;\mod\; m \Leftrightarrow \frac{x-y}{m}=k[/imath] for some integer [imath]k[/imath], but [imath]x[/imath], [imath]y[/imath] and [imath]m[/imath] don't have to be integer.