On primitive solutions of the Diophantine equation x² + y² = M

Lemma 1

Let a , b , a ˜ , b ˜ be integers and let M , N be positive integers such that ( a , b ) M and ( a ˜ , b ˜ ) N . Then

Proof

We have to verify that ( a a ˜ − b b ˜ ) 2 + ( a b ˜ + b a ˜ ) 2 = M ⋅ N . Indeed, we have

Fact 2

Let a , b , a ˜ , b ˜ be integers and let M , N be positive integers such that ( a , b ) M and ( a ˜ , b ˜ ) N . Then, for z ≔ ( a + i b ) ( a ˜ + i b ˜ ) , we have

Proposition 3

Let p = α α ¯ be a Pythagorean prime and let k be a positive integer. Then { ∣ Re ( α k ) ∣ , ∣ Im ( α k ) ∣ } is the unique positive, primitive solution to x 2 + y 2 = p k .

Proof

At first, we will show the existence of a primitive solution to the above equation. By observing that p k = α k α ¯ k , we see that this equation is satisfied by { ∣ Re ( α k ) ∣ , ∣ Im ( α k ) ∣ } . Thus, we need to show that these numbers are relatively prime. Assume not then there exist integers u , v , λ where λ > 1 such that α k = λ ( u + i v ) . By the uniqueness of prime factorisation in Z [ i ] we get λ = α l for some positive integer l . For θ = arg ( α ) , we get that θ π and tan ( θ ) are both rational, a contradiction to Niven’s theorem (see [23, Cor. 3.12]).

To show uniqueness, let us assume that a , b ∈ Z are relatively prime and satisfy a 2 + b 2 = p k . Then we also have

which implies that α divides either one of the factors on the left-hand side. Hence, without loss of generality, we have that α divides ( a + b i ) and by complex conjugation, α ¯ divides ( a − b i ) . However, neither α divides ( a − b i ) nor α ¯ divides ( a + b i ) . Otherwise, α or α ¯ would divide a and b (observe that α and α ¯ do not divide 2 because p is a Pythagorean prime), so both would then divide both because a , b ∈ Z . By considering α ≠ α ¯ , we get that p divides a and b , which is a contradiction to the coprimality of them. Hence, we conclude that α k divides ( a + b i ) and α ¯ k divides ( a − b i ) in the Gaussian integers. Therefore, both Gaussian integers on the left-hand side of the equation

are units, which implies the existence of s ∈ { 0 , 1 , 2 , 3 } such that i s α k = a + b i and we get our desired result

Theorem 4

Let n and k l be positive integers, p l = α l α l ¯ be pairwise distinct Pythagorean primes for 1 ≤ l ≤ n and let M = ∏ l = 1 n p l k l . Then

is a primitive solution for x 2 + y 2 = M .

Proof

Obviously, we have M = ∏ l = 1 n α l k l ∏ l = 1 n α l k l ¯ . Therefore, x 2 + y 2 = M is clearly satisfied by ∣ Re ( ∏ l = 1 n α l k l ) ∣ , ∣ Im ( ∏ l = 1 n α l k l ) ∣ .

It remains to show that our solution is relatively prime. If not, then there exists integers u , v , λ where λ > 1 such that ∏ l = 1 n α l k l = λ ( u + i v ) . In this case, we must have λ = ∏ l = 1 n α l k l ′ with 0 ≤ k l ′ ≤ k l . Additionally, it holds true that λ = λ ¯ = ∏ l = 1 n α l ¯ k l ′ . Observe that all prime factors of λ are different from ± 1 , ± i . Thus, we have a contradiction to the unique prime factorisation in Z [ i ] .□

Proposition 5

Let an arbitrary M = 2 r ∏ l = 1 n p l k l ∈ N > 2 be factorised and r , n ∈ N . If all the primes p l ≠ 2 are Pythagorean and r ∈ { 0 , 1 } , then there are 2 n − 1 positive, primitive solutions to x 2 + y 2 = M . Otherwise, there is no primitive solution.

Proof

At first assume that all the p l ’s are Pythagorean primes and r ∈ { 0 , 1 } . Let I , I ′ be a partition of the set { 1 , 2 , … , n } and

be factorised in Z [ i ] . Then each I gives us a primitive solution of M = Re ( α I ) 2 + Im ( α I ) 2 for r = 0 by Theorem 4. If r = 1 , then the solution clearly remains primitive.

Conversely, if { x , y } is a primitive solution to the equation x 2 + y 2 = M , then M = ( x + i y ) ( x − i y ) . So, both of these factors can be factorised by the Gaussian primes of M multiplied by a unit of Z [ i ] . Since these factorisations must be the complex conjugates of each other and ( x , y ) = 1 , there exists I ⊂ { 1 , 2 , … , n } and k ∈ { 0 , 1 , 2 , 3 } such that x + i y = ( 1 + i ) r i k α I or x + i y = ( 1 + i ) r ¯ i k α I . This shows that each primitive solution to the equation above can be constructed by the right choice of I .

Now we would like to show that x 2 + y 2 = M has exactly 2 n − 1 positive solutions. Let I 1 and I 2 be subsets of { 1 , 2 , … , n } and assume that α I 1 and α I 2 represent the same solution, i.e., we have

Then we find k ∈ { 0 , 1 , 2 , 3 } such that

Since i is a unit, we have that either both α I 1 and α I 2 have the same prime factors in Z [ i ] or they are the complex conjugates of each other. Therefore, k = 0 and α I 1 = α I 2 or α I 1 = α I 2 ¯ by definition of α I . Furthermore, if I 1 and I 2 are disjoint or equal, then ( ∗ ∗ ) is clearly satisfied, so we get the same positive, primitive solution. Thus, there are exactly 2 n − 1 different choices for I such that the resulting positive, primitive solutions are different from each other if all the p l ’s are Pythagorean primes and r ∈ { 0 , 1 } .

It remains to show the case where one of the primes p l is odd and non-Pythagorean or r > 1 . Let x , y ∈ Z with x 2 + y 2 = M .

Assume that p 1 ≡ 3 ( mod 4 ) . Then p 1 is a Gaussian prime and, without loss of generality, p 1 divides x + i y in Z [ i ] . Then p 1 also divides x − i y because its complex conjugate (which is p 1 itself) must divide the complex conjugate of x + i y . Hence, p 1 is also a divisor of the sum and the difference of both terms above. Since p 1 is not a divisor of 2, we get that p 1 divides x and y which let us conclude that our solution cannot be primitive.

Finally, we only have to treat the case r > 1 . Since 2 can be decomposed by the Gaussian primes 1 + i and 1 − i , we have that x + i y or x − i y must be divisible by the multiplication of at least two of these factors. Hence, x + i y and x − i y have a divisor 2 or 2 i . Thus, by similar arguments to above, you can show that x and y can be divided by 2, which shows us again that our solution cannot be primitive.

Alternative for r > 1 : (Shorter but not argued by the Gaussian prime theorem) If M is divisible by 4, then either x 2 ≡ 3 ( mod 4 ) and y 2 ≡ 1 ( mod 4 ) or the other way round because our solution is primitive, which implies that x and y must be odd. However, this is a contradiction because an integer square cannot be congruent to 3 ( mod 4 ) .□