2022/069) find the pythegorean triplet whose one number 10x+ y and largest number 10y + x

let the 3rd number be t

so $t^2 + (10x+y)^2 = (10y+x)^21$

or $t^2 = (10y+x)^2 - (10y+x)^2 = 99(y^2 - x^2) = 99(y-x)(y+x)= 11 . 3^2(y+x)(y-x)$

y+x is less than 18 and y-x is less than 10 

for the RHS to be a perfect square we must have y +x = 11 and y-x is to be perfect squiare and odd

because 11 cannot be a factor of y-x and if y +x is odd then y-x has to be odd because y-x = y + x - 2x

so y+ x = 11 and y-x = 1 giving y = 6 and x = 5

hence $t^2 = 99 * 11 * 1$ or t = 33

so triple is (33,56,65)