The sum of the squares of the digits of a positive​ two-digit number is 89. The difference between the two digits is 3. Find all​ two-digit numbers that satisfy this statement.

Answer 1

guessing...

30 does not work: sum of squares of the digit is 9

14 and 41 do not work, sum of squares of the digit is 17

25 and 52 do not work, sum of squares of the digit is 29

36 and 63 do not work, sum of squares of the digit is 45

47 and 74 do not work, sum of squares of the digit is 65

58 and 85 works: 25 + 64 = 89 !!!!

69 and 96 do not work, sum of squares of the digit is 117

limit reached...

apparently the answer is 58 and 85

Here's the alegraic solution

--------------------------------

t is the tens digit

u is the units digit

the number is 10t + u

t^2 + u^2 = 89

and either t-u = 3 or u-t = 3

supposing the former:

t-u=3 --> t = u+3

(u+3)^2 + u^2 = 89

u^2 + 6u + 9 + u^2 = 89

2u^2 + 6u + 9 = 89

2u^2 + 6u - 80 = 0

u^2 + 3u - 40 = 0

( u + 8 )( u - 5 ) = 0

u+8 results in negative measures

u-5 = 0 ---> u = 5

which forces t=8

so the number is 85

the digits differ by 3 and 64+25 = 89

and in the latter case:

then u-t = 3 ---> u = t+3

u-3 = t

(u-3)^2 + u^2 = 89

u^2 - 6u + 9 + u^2 = 89

2u^2 - 6u + 8 = 89

2u^2 - 6u - 80 = 0

u^2 - 3u - 40 = 0

(u - 8)(u+5) = 0

u+5 = 0 results in negative measures

u-8 = 0 ---> u = 8

t = u-3 = 8-3 = 5

so 58, as expected