Let a,b∈ℕ.If there exist x1,x2

實例：x1＝3，x2＝4，x3＝5，x4＝12；

y1＝6，y2＝8，y3＝9，y4＝12；z1＝87，z2＝116，z3＝123，z4＝164。因為 (3²＋4²＋5²＋12²)(6²＋8²＋9²＋12²)＝87²＋116²＋123²＋164²所以，(x1²＋x2²＋x3²＋x4²)(y1²＋y2²＋y3²＋y4²)＝z1²＋z2²＋z3²＋z4²

Proof :Let a1²＝x1²＋x2², a2²＝x3²＋x4², b1²＝y1²＋y2², b2²＝y3²＋y4²,

so, ab＝(x1²＋x2²＋x3²＋x4²)(y1²＋y2²＋y3²＋y4²)＝(a1²＋a2²)(b1²＋b2²)＝a1²b1²＋a1²b2²＋a2²b1²＋a2²b2²＝a1²b1²＋2a1a2b1b2＋a2²b2²＋a1²b2²－2a1a2b1b2＋a2²b1²＝(a1b1＋a2b2)²＋(a1b2－a2b1)²

As there exist (a1b1＋a2b2)²＝z1²＋z2² and (a1b2－a2b1)²＝z3²＋z4²,so, ab＝z1²＋z2²＋z3²＋z4².