Rational Solution

Write $r$ as a fraction $\frac {b}{c}$ :

$\sqrt{7-r^2} = \sqrt{7-\frac{b^2}{c^2}} = \frac{\sqrt{7c^2-b^2}}{c}$

This is a rational number for nonzero integers $b, c$ exactly when $7c^2-b^2$ is a perfect square.

Rewrite this as:

$a^2=7c^2-b^2 \\ a^2+b^2=7c^2$

Since the right side is divisible by 7, we have that:

$a^2 \equiv -b^2\ \pmod 7$

We can use the fact that 7 is prime to construct the inverse of the nonzero integer b,

$b^6 \equiv 1\ \pmod 7 \\ b^5 * b \equiv 1\ \pmod 7$

(Using Fermat's Little Theorem)

This simplifies to:

$(\frac{a}{b})^2 \equiv -1\ \pmod 7$

(where $\frac{a}{b}$ stands for $ab^{-1}$ , or $ab^5$ ).

Since the only square residues mod 7 are 0, 1, 2 and 4, and none of them are congruent to -1, there must be no integer solutions, so the initial expression is never a rational number when applied to rational arguments.

Suppose $r = \frac{m}{n}$ is a rational solution in reduced form, so that $m,n$ are coprime. Then $\sqrt{7n^2 - m^2} = n \sqrt{7-r^2}$ is rational, and therefore $7n^2 - m^2$ is an integer which is the square of a rational, hence it must be the square of an integer $k$ , so $8n^2 = n^2 + m^2 + k^2$ Considering this equation modulo $4$ and noting that integer squares are congruent to either $0$ or $1$ modulo $4$ , we have $n^2 + m^2 + k^2 \equiv 0 \pmod{4} \qquad \Rightarrow \qquad n^2 \equiv m^2 \equiv 0 \pmod{4}$ which implies that both $m,n$ are even, contradicting our assumption that they were coprime.

Since the only assumption was that $r$ was a solution, we conclude there are no solutions.

We are looking for rationals $p,q$ such that $p^2 + q^2 = 7$ . Finding a common denominator for these rationals, we are looking for integers $a,b,c$ with $c \neq 0$ such that $a^2 + b^2 = 7c^2$ .

Suppose that $a,b,c$ are integers with $a^2 + b^2 = 7c^2$ . Then $7$ is irreducible in the Gaussian integers $\mathbb{Z}[i]$ , and $7$ divides $(a+ib)(a-ib)$ . Thus $7$ must divide at least one of these factors in $\mathbb{Z}[i]$ , and hence divides both $a$ and $b$ in $\mathbb{Z}$ . Thus $a = 7A$ , $b=7B$ for integers $A,B$ , and hence $7(A^2 + B^2) = c^2$ , so that $7$ divides $c$ , and hence $c=7C$ where $C$ is an integer with $A^2 + B^2 = 7C^2$ . Since we can continue dividing by $7$ in this way, an infinite descent argument tells us that the only solution is $a=b=c=0$ . Thus there are no solutions with nonzero $c$ , and hence there are no rationals $p,q$ with $p^2+q^2=7$ .

Let r = a/b. substituting and squaring, we get the following expression: 7k^2 = A^2 +B^2, for integers A,B, and k. But the sum of two squares is not divisible by 7. Hence, no solutions. Ed Gray