One positive integer is 3 3 less than twice another. The sum of their squares is 954 954 . Find the integers.

Question

One positive integer is 3 less than twice another.  The sum of their squares is 954.  Find the integers.

Answer 1

Let one integer be x then the other is 2x-3.

Sum of their squares is 954, so x²+(2x-3)²=954.

Therefore, x²+4x²-12x+9=954, 5x²-12x-945=0.

945=3³×5×7, so we can work out the factors in pairs:

(3,315), (5,189), (7,135), (9,105), (15,63), (21,45), (27,35).

We can see that 5×15-63=12, so we can factorise: (5x+63)(x-15)=0. The only positive zero is x=15, so the integers are 15 and 2x-3=27.

Check: 15²+27²=225+729=954. OK.