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Answer 1

Step-by-step explanation:

Let a = 10x + y and b = 10y + x

$\bf{a}^{2} - {b}^{2} = {k}^{2} \\ \bf \: (10x + y)^{2} - (10x - x)^{2} = {k}^{2}$

Now expanding

$\bf \: 100 {x}^{2} + 20xy + {y}^{2} - 100 {y }^{2} - 20xy - {y}^{2} = {k}^{2} \\ \bf \: 99 {x}^{2} - 99 {y}^{2} = {k}^{2} \\ \bf \: 99( {x}^{2} - {y}^{2} ) = {k}^{2} \\ \bf \: 99 [x + y][x - y] =  {k}^{2}$

Since , k is an integer k^2 is a perfect square.

Case 1

$\bf \: 99 \times 99 = {k}^{2} \: \\ \bf Therefore \: [x + y] \: [x - y] \: = 99, \\ \bf \: which \: gives \: us \: that \: x + y \: = 11 \: and x - y = 9$

Case 2

$\bf \: 99 can \: be \: written \: as \:  11 \times {3}^{2} , \\ \bf \: which \: means \: to \: make \: the \: expression \: a \: perfect square, \: \\ \bf [x + y][x - y] = 11$$\bf \: This \: gives \: us \: x + y = 11 and \: x - y = 1 \\ \bf \: The \: values \: that \: satisfies \: these \: equations \: would \: be \: \\ \bf x = 6 \: and y = 5. \: \\ \bf These \: satisfy \:  a and b \:  (65 \: and \: 56). \\ \bf \: So \: k \: = 11 \times 3 = 33 \\ \bf \: a + b + k = 10x + y + 10y + x + k \\ \bf= 11(x + y) + 33 = 11(11) + 33 = 154$