Number Theory Assignment II

Miliyon T.Addis Ababa University

Department of Mathematics

March 1, 2015

Let p be a prime of the form p 1(mod4) and let (x, y, z) = 1 be a primitive solution tothe Diophantine equation

x2 + py2 = z2 (1)

1. Show that z is necessarily odd.

Solution: Suppose z is even.If y is even py2 is even, then x2 is even so is x. But this implies (x, y, z) 2 whichcontradict the fact that (x, y, z) = 1.If x is even x2 is even, then py2 is even since p is odd [ p 6= 2 as p is of the formp 1( mod 4)]. y2 is even so is y. Thus (x, y, z) 2 which contradict the factthat (x, y, z) = 1.Let both x and y be odd. Thus x2 1( mod 4) and y2 1( mod 4) thusx2 + py2 2( mod 4) which is impossible since the square of every even integeris congruent to 0 modulo 4. In any case we arrived at a contradiction. Hence ourassumption that z is even is false. z is odd.

2. Assume that y is even and characterize the solution under different cases p|(z + x) orp|(z x).Solution: Suppose y is even. From #1 we know that z is odd. So z2 is odd. Since y is

even y2 is even clearly py2 is even. So x2 should be odd but in order to have thatx must be odd. Thus the Diophantine equation x2 + py2 = z2 can be rewritten as

py2 = (z x)(z + x) (2)

Since both z and x are odd we have (z x) and (z + x) both even.So (z x) = 2a a = (zx)

2and (z + x) = 2b b = (z+x)

2for a, b Z. Then (2)

1

becomes

py2 = (2a)(2b)

= 4ab

p(y

2

)2= ab as y is even.

[Case 1:] If p|(z + x)(

y2

)2=

(ap

)b If we assume p|(z + x) then p|2a but p - 2 as p 1( mod 4).

Thus p|a.Let d =

(ap, b

) d|

(ap

)and d|b d|

((zx)

2

)and d|

((z+x)

2

)This implies

d|( (zx)2

+ (z+x)2

) d|z ,and d|( (zx)2 (z+x)

2) d|x

Thus d = 1 as (x, z) = 1. Hence (ap, b) = 1.

Now we have (y2)2 = (a

p)b with (a

p, b) = 1.

The product of two co-prime integers (ap) and b is a square number. Thus (a

p)

and b are square numbers. So there are integers m and n such that (ap) = m2 and

b = n2, (m,n) = 1 as (ap, b) = 1.

So (y2)2 = (a

p)b y = 2mn and (z+x)

2= pm2 and (zx)

2= n2

Thus z + x = 2pm2 and z x = 2n2Finally we get the solution

z = n2 + pm2, y = 2mn, x = pm2 n2 with (m,n) = 1 and m&n are opposite in parity.To show m&n are opposite in parity: suppose both m&n are even. This impliesm2 and n2 are even but this shows us (a

p, b) 2 which contradicts the fact that

(ap, b) = 1.

Suppose both m&n are odd. Thus z = n2 + pm2 and x = pm2 n2 are even. So(x, z) 2 but this contradicts the fact that z and x are relatively prime.

[Case 2:] If p|(z x)In a similar manner we will get a solution

z = m2 + pn2, y = 2nm, x = m2 pn2 with (m,n) = 1 and m&n are opposite in parity.3. Assume that y is odd and characterize the solution under different cases p|(z + x) or

p|(z x).Solution: Suppose y is odd. So, py2 is odd.

Now x2 + py2 = z2 py2 = (z x)(z + x). Thus both (z x) and (z + x) areodd as py2 is. Since (x, z) = 1 If we let d = (z x, z + x), then either d = 1 ord = 2. But d 6= 2 because both (z x) and (z + x) are odd. Hence d = 1.

[Case 1:] If p|(z + x)

y2 = ((z + x)/p)(z x) with ((z + x)p

, z x) = 1.

2

Similarly as we have done above there exist integers m and n such that (z+x)p

= m2

and z x = n2 with (m,n) = 1. Since y is odd both m and n are odd.This gives us a solution

z = (n2 + pm2)/2, y = mn, x = (pm2 n2)/2 with (m,n) = 1 and m,n odd integers.

[Case 2:] If p|(z x)In a similar manner we will get a solution

z = (m2 + pn2)/2, y = mn, x = (m2 pn2)/2 with (m,n) = 1 and m,n odd integers.

4. Use the above result to give a concrete solution to the Diophantine equation x2+5y2 =z2 when x is even and x is odd.

Solution: (1) Suppose x is even.

x2 + 5y2 = z2 5y2 = (z x)(z + x)

We know that z is odd and 5 is a prime of the form 5 1( mod 4). Thus both(z x) and (z + x) are odd. So, 5y2 is odd which implies y is odd. Now thequestion resembles #3. So,

[Case 1:] If 5|(z + x), then the solution

z = (n2 + 5m2)/2, y = mn, x = (5m2 n2)/2 with (m,n) = 1 and m,n odd integers.

[Case 2:] If 5|(z x), then the solution

z = (m2 + 5n2)/2, y = mn, x = (m2 5n2)/2 with (m,n) = 1 and m,n odd integers.

(2) Suppose x is odd.

x2 + 5y2 = z2 5y2 = (z x)(z + x)

We know that z is odd and 5 is a prime of the form 5 1( mod 4). Thus both(z x) and (z + x) are even. So, 5y2 is even which implies y is even. Now thequestion resembles #2. So,

[Case 1:] If 5|(z + x), then the solution becomes

z = n2 + 5m2, y = 2mn, x = 5m2 n2 with (m,n) = 1 and m&n are opposite in parity.

[Case 2:] If 5|(z x), then the solution becomes

z = m2 + 5n2, y = 2nm, x = m2 5n2 with (m,n) = 1 and m&n are opposite in parity.

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