Problems & Puzzles: Puzzles

Puzzle 968. Another property of primes 4m+1.

Sebastián Martín Ruiz made the following empirical discovery:

"when  p  is a prime of the form  4m + 1, the number of pairs  (a,b)  (with 1 <= a <= b <= p)  for which  a^2 + b^2  is a multiple of  p, equals  p."

Examples given by Sebastián:

For p=5, there are 5 of such pairs:


For p=13, there are 13 of such pairs:

As a matter of fact Sebastián tested his observation for all prime 4m+1 <= 509.

Another friend of these pages verified the same observation for all prime 4m+1 <= 6301.

Q. Can you find a proof of this empirical statement?


Contributions came from Adam Stinchcombe, Emmanuel Vantieghem, Ken Wilke, Oscar Volpatti.


Adam wrote on Aug 31, 2019:

Euler (and others) proved Fermat's claim that a prime of the form 4m+1 can be expressed as the sum of two squares, p=x^2+y^2.  Then (x/y)^2==-1 mod p is a square root of -1, call it w: w==x/y mod p, w^2==-1 mod p.  For each a, a^2 + (wa)^2 == a^2+w^2a^2 == a^2+(-1)a^2==0 mod p.  So for each a, 1<=a<p, there exist (at least one) b (equal to w*a mod p) with a^2+b^2==0 mod p.

Suppose (a,b1) and (a,b2) are two solutions to x^2+y^2==0 mod p with 1<=a,b1,b2<p.  Then a^2+b1^2==0==a^2+b2^2 mod p so b1^2==b2^2 and (b1/b2)^2==1 and b1/b2==1 or -1 mod p.  Either b1==b2 or b1==-b2==p-b1 mod p.  Only one of these numbers is b1,b2 is less than p/2.  Similarly, for the first coordinate a=a1=a2.  So. for a given solution (a,b), there is a single (a',b') amongst the four {(a,b),(a,p-b),(p-a,b),(p-a,p-b)} with a' and b' both less than p/2.

There are (p-1)/2 residues >=1 and <p/2, these pair off into (p-1)/4 pairs that solve x^2+y^2==1 mod p.  Each pair generates the four solutions {(a,b),(a,p-b),(p-a,b),(p-a,p-b)}.  This results in 4*(p-1)/4 = p-1 solutions.

Then there is the solution (p,p), giving a total of p solutions.


Emmanuel wrote on Set 1, 2019:

To every solution  (a,b)  of our problem corresponds a solution of the congruence
     x^2 + y^2 == 0 (mod p).
but not conversely since we added the condition  a <= b ((the solution  (x,y)  is considered different from  (y,x), except when  x == y == 0)
It is easy to show that the congruence has exactly  2p-1  solutions.
Indeed, if  p  is of the form  4m+1, there exist an integer  w  such that  w^2 == -1 (mod p).
Thus, the equation (*)  can be written as  (x-wy)(x+wy) == 0 (mod p).
So, the solutions are  (0,0), (wt,t), (-wt,t)  with  t = 1, 2, ..., p-1.
Thus, the number of solutions of our problem is exactly  p, Q.E.D.


Ken wrote on Set 2, 2019:

Proof:  Let p be a prime of the form 4m+1 for some integer m. Clearly if (a, b) = (p, p) then  a^2 +b^2==0 (mod p). If (1 <= a <= b < p) , then a^2 +b^2==0 (mod p) and         (a*b’)^2 + 1 ==0 (mod p) where b’ is the unique integer such that b*b’==1 (mod p).

By [1], it is well known that congruence  x^2 + 1 ==0 (mod p) (1)

has the solution  x == + ((p-1)/2))! (mod p).

Let 1, 2,. . ., p be a complete set of residues (mod p). Let g denote ((p-1)/2))! (mod p). Then g^2 + 1 ==0 (mod p) (2)

Multiplying the congruence (2) term wise by the residues 1, 2,. . ., p-1,p we generate p pairs of integers of the form (ci, di) where (ci)^2 +(di)^2 == (g*i)^2 +(i)^2 ==0 (mod p) for each residue I = 1,2,   ,p. Note that if one uses the negative value of x, this merely changes the sign of (g*i) and does not otherwise affect the values of the congruences.

(ci)^2 +(di)^2 == (g*i)^2 +(i)^2 ==0 (mod p)

Finally looking at the list of pairs of integers (ci, di) by taking ai as the smallest value in the pair(ci, di) and bi as the largest value in the pair (ci, di), one easily generates the values of (a,b) discovered by the proposer.

Nagell, Introduction to Number Theory, Chelseas Publishing Company, New York, Second Edition, (1964), pp.99-100, Theorem 58.n 


Oscar wrote on Set 4, 2019

Sebastián's statement is true.
I'll use the following notations:
"y = x % p"  means  "y is the remainder after division of x by p, with 0 <= y <= p-1";
"y == x  mod p" means "y is congruent with x modulo p".
Choose a prime p of the form 4m+1.
Lagrange proved the following lemma: p divides the number q = 1+r^2, with r = ((2m)!) % p.
Apply Wilson's theorem to the prime p = 4m+1:
(4m)! == -1  mod p.
Break the left factorial into two products:
(1)*(2)*...*(2m) == r  mod p, by definition of r;
(2m+1)*...*(4m-1)*(4m) == (-2m)*...*(-2)*(-1) == r*(-1)^(2m) == r  mod p.
Therefore, r^2 == -1  mod p, or equivalently  1+r^2 == 0  mod p, as claimed by Lagrange.
Clearly, p doesn't divide r, otherwise it would divide the difference q - r^2 = 1.
Lagrange's lemma helps us to find all the integer pairs (a,b) for which a^2 + b^2 is a multiple of p.
a^2 + b^2 == 0  mod p;
b^2 == (-1)*(a^2) == (r^2)*(a^2) == (ra)^2  mod p;
0 == b^2 - (ra)^2 == (b-ra)(b+ra)  mod p.
If the prime p divides the product of two or more factors, then it divides at least one of the factors.
Therefore, b == ra  mod p, or b == -ra  mod p.
Such congruences simultaneously hold when p divides 2ra; in this case, as p is odd and coprime with r, it divides a (and it divides b too).
The square 1 <= a,b <= p contains 2p-1 pairs (a,b) for which a^2 + b^2 is a multiple of p.
Check, row by row, the congruences previously found:
for 1 <= a <= p-1, we obtain two pairs (a, (ra) % p) and (a, (-ra) % p);
for a = p, we obtain one pair (p,p).
This set of solutions is necessarily symmetric around the axis b = a: if it contains the pair (x,y), it must contain the pair (y,x) too.
The pair (p,p) is the only solution along the axis b = a: if p divides 2*a^2, then it must divide a.
So, the inequality a < b must hold for exactly half of the remaining 2p-2 solutions.
Therefore, the triangle 1 <= a <= b <= p contains 1+(p-1) = p solutions, as claimed by Sebastián


Records   |  Conjectures  |  Problems  |  Puzzles