AoPS - Problem

Let in interval fourth powers $a^4,...(a+9)^4$ and cubes $b^3,...(b+99)^3$. $(b+99)^3\ge 1000000\to (a+10)^>10^6\to a\ge 22\to b\ge 61\to (a+10)^4>160^3\to a\ge 35\to b\ge 114\to (a+10)^4> 213^3\to a\ge 46\to .b\ge 164\to a\ge 56\to b\ge 214\to (a+10)^4>313^3\to a\ge a\ge 65\to b\ge 261\to (a+10)^4>360^3\to a\ge 73\to b\ge 305\to (a+10^4>404^3\to a\ge 81\to b\ge 350\to (a+10)^4\ge 449^3\to a\ge 88\to b\ge 391\to (a+10)^4>490^3\to a\ge 95\to b\ge 433\to (a+10)^4>532^3\to a\ge 101\to b\ge 470\to (a+10)^3\ge 569^3\to a\ge 107\to b\ge 507\to a\ge 113.$ It is enaff to prove, that sequence had at least $122^2-113^2+1=2116$ perfect squares. If we continued calculathion... $a\ge 200\to b\ge 1169\to (a+10)^4>1268^3\to a\ge 203 ... \to a\ge 400\to a\ge 401$. Stabilished, when a about 500. It give about 9000 perfect squares.