Infinite Descent I

If \(a\) is odd, then \(a=2k+1\). Then \(a^2=4k^2+4k+1\), which is odd. Hence if \(a^2\) is even, \(a\) cannot be odd, so \(a\) is even.

Since \(a^2=a\cdot a\) and the prime \(p\) divides this product, Euclid's lemma gives that \(p\) divides at least one factor. Both factors are \(a\), so \(p\mid a\).

If \(x^2=2y^2\), then \(x^2\) is even, so \(x=2u\). Then \(4u^2=2y^2\), hence \(y^2=2u^2\), and \(y\) is also even: \(y=2v\). Then \(u^2=2v^2\), a smaller positive solution. Infinite descent is impossible.

Let \(\sqrt{2}=\frac{a}{b}\), where \(\gcd(a,b)=1\). Then \(a^2=2b^2\). Hence \(a\) is even, \(a=2c\). Then \(4c^2=2b^2\), so \(b^2=2c^2\), and \(b\) is even. Thus \(a\) and \(b\) have common divisor \(2\), a contradiction.

From \(x^2=3y^2\), we get \(3\mid x^2\), so \(x=3u\). Then \(9u^2=3y^2\), hence \(y^2=3u^2\). Therefore \(3\mid y\). Dividing both \(x\) and \(y\) by \(3\) gives a smaller positive solution, impossible by descent.

Let \(\sqrt{3}=\frac{a}{b}\), \(\gcd(a,b)=1\). Then \(a^2=3b^2\), so \(3\mid a\), \(a=3c\). We get \(9c^2=3b^2\), hence \(b^2=3c^2\), and \(3\mid b\). This contradicts \(\gcd(a,b)=1\).

Let \(\sqrt{5}=\frac{a}{b}\) in lowest terms. Then \(a^2=5b^2\), so \(5\mid a\). Let \(a=5c\). Then \(25c^2=5b^2\), hence \(b^2=5c^2\), and \(5\mid b\). This contradicts lowest terms.

From \(x^2=8y^2\), \(x\) is even. Let \(x=2u\). Then \(4u^2=8y^2\), so \(u^2=2y^2\). But this equation has no positive integer solutions. Hence the original equation has none.

From \(x^2=12y^2\), \(x\) is even: \(x=2u\). Then \(4u^2=12y^2\), so \(u^2=3y^2\). But \(u^2=3y^2\) has no positive integer solutions by descent with prime \(3\). Contradiction.

Squares modulo \(3\) are only \(0\) and \(1\). For the sum of two such residues to be divisible by \(3\), the only possibility is \(0+0\). Thus \(x^2\equiv0\pmod3\) and \(y^2\equiv0\pmod3\). Therefore \(3\mid x\) and \(3\mid y\).

From \(3\mid x^2+y^2\), we get \(3\mid x\) and \(3\mid y\). Let \(x=3x_1\), \(y=3y_1\). Then \(9x_1^2+9y_1^2=3z^2\), hence \(z^2=3x_1^2+3y_1^2\), so \(3\mid z\). Dividing \(x,y,z\) by \(3\) gives a smaller positive solution. Infinite descent is impossible.

Squares modulo \(5\) are \(0,1,4\). Thus \(2y^2\) can have residues \(0,2,3\). The sum \(x^2+2y^2\) can be \(0\) modulo \(5\) only as \(0+0\). Therefore \(x^2\equiv0\) and \(y^2\equiv0\pmod5\), hence \(5\mid x\), \(5\mid y\).

From \(5\mid x^2+2y^2\), we get \(5\mid x\), \(5\mid y\). Let \(x=5x_1\), \(y=5y_1\). Then \(25x_1^2+50y_1^2=5z^2\), so \(z^2=5x_1^2+10y_1^2\), hence \(5\mid z\). Dividing all variables by \(5\) gives a smaller positive solution. Descent is impossible.

The quadratic residues modulo \(7\) are \(0,1,2,4\). No two nonzero elements of \(\{1,2,4\}\) sum to \(0\) modulo \(7\). Hence \(x^2+y^2\equiv0\pmod7\) is possible only when \(x^2\equiv y^2\equiv0\pmod7\). Therefore \(7\mid x\), \(7\mid y\).

The equation implies \(7\mid x^2+y^2\), so \(7\mid x\) and \(7\mid y\). Let \(x=7x_1\), \(y=7y_1\). Then \(49x_1^2+49y_1^2=7z^2\), hence \(z^2=7x_1^2+7y_1^2\), so \(7\mid z\). Dividing all variables by \(7\) gives a smaller solution of the same type. Infinite descent is impossible.

Let \(\sqrt{6}=\frac{a}{b}\) in lowest terms. Then \(a^2=6b^2\), so \(a\) is even: \(a=2c\). We get \(4c^2=6b^2\), so \(2c^2=3b^2\). The right-hand side is even, hence \(b^2\) is even, and \(b\) is even. This contradicts lowest terms.

From \(x^2=2y^2\), \(x\) is even, \(x=2u\). Then \(y^2=2u^2\), so \(y\) is even. Thus \(x\) and \(y\) have common divisor \(2\), contradicting coprimality.

Since \(3\mid x^2\), we have \(x=3u\). Then \(9u^2=45y^2\), hence \(u^2=5y^2\). But \(u^2=5y^2\) has no positive integer solutions: it forces both \(u\) and \(y\) to be divisible by \(5\), starting descent. Contradiction.

From \(a^2=2b^2\), both \(a\) and \(b\) are even. Then \(a=2a_1\), \(b=2b_1\), and \(\left(\frac{a_1}{b_1}\right)^2=2\). But \(b_1=\frac{b}{2}

Suppose \(3\mid x^2+y^2\). Then modulo \(3\), both squares must be \(0\), so \(3\mid x\) and \(3\mid y\). This contradicts \(\gcd(x,y)=1\). Therefore \(x^2+y^2\) is not divisible by \(3\).

Assume a solution exists and choose one with minimal \(x+y\). If \(x=y\), then \(2x^2=3x^2\), impossible. Let \(x>y\). The equation \(x^2-3xy+y^2=0\), viewed as a quadratic in \(x\), has second root \(x'=3y-x\). By Vieta, \(xx'=y^2\), so \(x'>0\), and since \(x>y\), \(x'=\frac{y^2}{x}

As a quadratic in \(x\), the equation is \(x^2-4yx+y^2=0\). Its discriminant must be a square: \(D=16y^2-4y^2=12y^2\). Thus there is an integer \(u\) such that \(u^2=12y^2\). But this equation has no positive integer solutions, as proved earlier. Contradiction.

Rewrite as \(x^2-5yx+2y^2=0\). The discriminant \(D=25y^2-8y^2=17y^2\) must be an integer square. This would mean \(u^2=17y^2\) for some positive \(u\). But \(u^2=17y^2\) is impossible by the same descent with prime \(17\). Contradiction.

Assume a solution exists and choose one with minimal sum \(x+y\). Let \(x\ge y\). If \(y=1\), then \(x^2+1=3x+2\), so \(x^2-3x-1=0\), which has no integer root.

Thus \(y\ge2\). View the equation as a quadratic in \(x\): \(x^2-3yx+(y^2-2)=0\). The second root \(x'=3y-x\) is an integer. By Vieta, \(xx'=y^2-2>0\), so \(x'>0\). If \(x=y\), then \(2y^2=3y^2+2\), impossible, hence \(x>y\). Therefore \(x'=\frac{y^2-2}{x}