## (1 + √2)^n = √x + √(x+1); for n = 1,2,3,…,20

$( \,1 + \sqrt {2} \,)^n \; = \; \sqrt { \,x \,} \; + \; \sqrt { \,x+1 \,}$

For   $n = 1, 2, 3, ..., 20$

$( \,1 + \sqrt {2} \,) \; = \; \sqrt { \,1 \,} \; + \; \sqrt { \,2 \,}$
$( \,1 + \sqrt {2} \,)^2 \; = \; \sqrt { \,8 \,} \; + \; \sqrt { \,9 \,}$
$( \,1 + \sqrt {2} \,)^3 \; = \; \sqrt { \,49 \,} \; + \; \sqrt { \,50 \,}$
$( \,1 + \sqrt {2} \,)^4 \; = \; \sqrt {288} \; + \; \sqrt {289}$
$( \,1 + \sqrt {2} \,)^5 \; = \; \sqrt {1681} \; + \; \sqrt {1682}$
$( \,1 + \sqrt {2} \,)^6 \; = \; \sqrt {9800} \; + \; \sqrt {9801}$
$( \,1 + \sqrt {2} \,)^7 \; = \; \sqrt {57121} \; + \; \sqrt {57122}$
$( \,1 + \sqrt {2} \,)^8 \; = \; \sqrt {332928} \; + \; \sqrt {332929}$
$( \,1 + \sqrt {2} \,)^9 \; = \; \sqrt {1940449} \; + \; \sqrt {1940450}$
$( \,1 + \sqrt {2} \,)^{10} \; = \; \sqrt {11309768} \; + \; \sqrt {11309769}$
$( \,1 + \sqrt {2} \, )^{11} \; = \; \sqrt {65918161} \; + \; \sqrt{65918162}$
$( \,1 + \sqrt {2} \, )^{12} \; = \; \sqrt {384199200} \; + \; \sqrt {384199201}$
$( \,1 + \sqrt {2} \, )^{13} \; = \; \sqrt {2239277041} \; + \; \sqrt {2239277042}$
$( \,1 + \sqrt {2} \, )^{14} \; = \; \sqrt {13051463048} \; + \; \sqrt {13051463049}$
$( \,1 + \sqrt {2} \, )^{15} \; = \; \sqrt {76069501249} \; + \; \sqrt {76069501250}$
$( \,1 + \sqrt {2} \, )^{16} \; = \; \sqrt {443365544448} \; + \; \sqrt {443365544449}$
$( \,1 + \sqrt {2} \, )^{17} \; = \; \sqrt {2584123765441} \; + \; \sqrt {2584123765442}$
$( \,1 + \sqrt {2} \, )^{18} \; = \; \sqrt {15061377048200} \; + \; \sqrt {15061377048201}$
$( \,1 + \sqrt {2} \, )^{19} \; = \; \sqrt {87784138523761} \; + \; \sqrt {87784138523762}$
$( \,1 + \sqrt {2} \, )^{20} \; = \; \sqrt {511643454094368} \; + \; \sqrt {511643454094369}$

A115598   Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.
1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, …

A115599   Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-X values.
2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, …

$1^3+2^3+3^3+...+x^3 = y^4 = (T(x))^2$

$1^3+2^3+3^3+...+8^3 = 6^4 = (T(8))^2$
$1^3+2^3+3^3+...+49^3 = 35^4 = (T(49))^2$
$1^3+2^3+3^3+...+288^3 = 204^4 = (T(288))^2$
$1^3+2^3+3^3+...+1681^3 = 1189^4 = (T(1681))^2$
$1^3+2^3+3^3+...+9800^3 = 6930^4 = (T(9800))^2$
$1^3+2^3+3^3+...+57121^3 = 40391^4 = (T(57121))^2$
$1^3+2^3+3^3+...+332928^3 = 235416^4 = (T(332928))^2$
$1^3+2^3+3^3+...+1940449^3 = 1372105^4 = (T(1940449))^2$
$1^3+2^3+3^3+...+11309768^3 = 7997214^4 = (T(11309768))^2$
$1^3+2^3+3^3+...+65918161^3 = 46611179^4 = (T(65918161))^2$
$1^3+2^3+3^3+...+384199200^3 = 271669860^4 = (T(384199200))^2$
$1^3+2^3+3^3+...+2239277041^3 = 1583407981^4 = (T(2239277041))^2$
$1^3+2^3+3^3+...+13051463048^3 = 9228778026^4 = (T(13051463048))^2$
$1^3+2^3+3^3+...+76069501249^3 = 53789260175^4 = (T(76069501249))^2$
$1^3+2^3+3^3+...+443365544448^3 = 313506783024^4 = (T(443365544448))^2$
$1^3+2^3+3^3+...+2584123765441^3 = 1827251437969^4 = (T(2584123765441))^2$
$1^3+2^3+3^3+...+15061377048200^3 = 10650001844790^4 = (T(15061377048200))^2$
$1^3+2^3+3^3+...+87784138523761^3 = 62072759630771^4 = (T(87784138523761))^2$
$1^3+2^3+3^3+...+511643454094368^3 = 361786555939836^4 = (T(511643454094368))^2$