Show that 7 is not the sum of 2 rationale squares?

Question

A response using congruences is preferred.

Answer 1

i know but i am not doing your home work.

Answer 2

a^2/b^2 + c^2/d^2 = 7

(a^2d^2 +b^2c^2)/(b^2d^2) = 7

a^2d^2 +b^2c^2 = 7b^2d^2

The left hand side clearly always even and the right hand side is clearly always odd. Therefore an odd number would be equal to an even number which is impossible. Thus 7 cannot be the sum of the squares of two rational numbers.  

Answer 3

so numbers have to be rational

let first number be a/b and second number be x/y where a,b,x and y are integers

(a/b)^2+(x/y)^2= 7

a^2/b^2 + x^2/y^2 = 7

a^2/b^2= 7-x^2/y^2

find LCD for right side

(a/b)^2 = (7y^2 -x^2) /y^2

now factor out right side

(a/b)^2 = (y√7 -x) (y√7 +x) /y^2

a/b = √(y√7 -x) (y√7 +x) / y

so b and y in denominator as integers are fine

but a = √(y√7 -x) (y√7 +x)  making a not a rational integer since it is the square root of irrational product

also it can't be a multiple of integer a either since the square roots would still make it rationalso it can't be of rational squares

hope this helps  

Answer 4

Suppose u² + v² = 7, where u and v are rational numbers.

By multiplying by the LCM of the denominators, it

can be reduced to solving

x² + y² = 7z², where

x, y , z are integers.

Also, by dividing out common factors we can assume

x, y, z are relatively prime in pairs.

First, suppose z is odd. Then z² = 1(mod 4), so

x² + y² = 3(mod 4), which is impossible.

If z is even, we get

x² + y² = 0(mod 4).

But this can happen only if both x and y are even,

which contradicts the fact that x and y are relatively prime.

Thus there are no solutions to the original equation.

Answer 5

Assume there exist rationals x and y such that x^2 + y^2 = 7.  This equation can be normalized to a^2 + b^2 = 7c^2, where a, b, c are integers, by multiplying through by the least common multiple of the denominators of x and y.  

We show by infinite descent that there can be no integral solution to the equation a^2 + b^2 = 7c^2.  By our initial assumption, there exists at least one solution to this equation, so by the well-ordering principle there must exist a smallest solution (let's say "smallest" means in terms of the quantity a^2 + b^2 + c^2).  Let (a, b, c) be that particular smallest solution.  Also let 0 < a, b, c for convenience (we can always do this).

The right hand side of a^2 + b^2 = 7c^2 is congruent to 0 mod 7, so the left hand side must be 0 as well.  But, since the squares mod 7 are 0, 1, 2, and 4, the only way to get 0 as a sum of two squares mod 7 is for both of the squares to be 0 themselves.  Thus, a and b are both congruent to 0 mod 7.

Let a = 7k, b = 7m.  Substituting gives 49k^2 + 49m^2 = 7c^2, or 7k^2 + 7m^2 = c^2.  The left hand side of this equation is congruent to 0 mod 7, so the right hand side must be 0 as well.  Therefore, c = 7n, and substituting once again gives 7k^2 + 7m^2 = 49n^2, or k^2 + m^2 = 7n^2.  

0<k<a, 0<m<b, 0<n<c, so k^2 + m^2 + n^2 < a^2 + b^2 + c^2.  But we concluded that k^2 + m^2 = 7n^2, which tells us that (k, m, n) is a SMALLER solution to the initial equation than (a, b, c) was.  But this is a contradiction, as (a, b, c) was supposed to be the smallest solution to the equation.  Therefore, such a smallest solution cannot exist, and so we can conclude that no integral solutions (a, b, c) exist to the equation a^2 + b^2 = 7c^2.  But this implies that 7 cannot be expressed as the sum of two rational squares, so we are done.

Just as a side note, ironduke's solution doesn't work, since the right hand side of his equation can easily be even (if either b or d is even).  Unfortunately, I don't know of a shorter solution than the one I gave you.