Problem
NT-B2-M08-P014 An Equation With \(11\)
#14
★★★★☆ Level 4 of 5
Prove that \(x^2+y^2=11z^2\) has no nonzero integer solutions.
Use the previous problem for \(p=11\).
Since \(11\equiv 3\pmod 4\) and \(11\mid x^2+y^2\), we get \(11\mid x\) and \(11\mid y\). Then \(121\mid x^2+y^2=11z^2\), so \(11\mid z\). Dividing all variables by \(11\), we obtain a smaller nonzero solution. Infinite descent is impossible.
A short application of the general lemma.