Couldn’t you imagine you are dealing with a triangle with a hypotenuse of 106. Make a small table of all integers and their squares and test that each pair of consecutive to see which pair satisfies the Pythagorean theorem. Does that help?

Wouldn’t that be geometry?

What I did to solve it was make a short list of odd numbers (1, 3, 5, 7, 9). From that point it was easy to see that:
1 and 5 couldn’t work – 1 times 1 is 1, 5 times 5 is 25, 1 plus 25 is 26.
3 and 7 couldn’t work – 3 times 3 is 9, 7 times 7 is 49, 9 plus 49 is 58.
Then 5 and 9, and they did work – 5 times 5 is 25, 9 times 9 is 81, 25 plus 81 is 106.

An odd integer is of the form 2n+1 for some integer n. Three consecutive odd integers are then 2n+1, 2n+3, and 2n+5. The sum of the squares of the first and third is (2n+1)^2+(2n+5)^2, so the equation is (2n+1)^2+(2n+5)^2 = 106. Simplified, this is n^2+3n-10 = 0 = (n+5)(n-2). The answers for n are -5 and 2, and the odd integers are (-9, -7 -5) or (5, 7, 9). (2*(-5) +1)^2 + (2*(-5)+5)^2 = (-9)^2+(-5)^2 = 81+25 = 106 and (2*2+1)^2+(2*2+5)^2 = 5^2+9^2 = 25+81 = 106.

Or…....
If x is one of the digits then the other is (x+4)
So the algebraic equation is x^2 + (x+4)^2 = 106
Therefore x is 5 ( or -9 if we include negative solutions)

@flutherother, thanks for making me look like an idiot—I didn’t see that the oddness was a red herring.

@ratboy No offence meant. There is more than one way to skin a cat.

Another way, solve using (x-2), x, (x+2) as three numbers. Then, x = +/- 7. Sum of 1st and 3rd terms = 106 for both positive and negative valus. So, three numbers can be either 5,7,9 or -5,-7,-9.

@flutherother, my remark was a compliment.

Thanks all for giving me some insight.