Question 1094813
<br>I don't see a workable algebraic solution....<br>
The condition that the squares of three integers a, b, and c are in arithmetic progression, with c being 12 greater than a will be satisfied if we
(1) choose an integer value for a;
(2) set c = a+12;
(3) evaluate a^2 and c^2; and
(4) find the average of a^2 and c^2<br>
If that average is a perfect square, then that average makes an arithmetic sequence with a^2 and c^2; then the second integer we are looking for is the square root of that average.<br>
So try a=1; if that doesn't give you a solution, try a=2; and so on.<br>
a = 1; c = 13; a^2 = 1; c^2 = 169; the average of a^2 and c^2 is 85... not a perfect square.<br>
a = 2; c = 14; a^2 = 4; c^2 = 196; the average of a^2 and c^2 is 100...AHA!<br>
The first and last integers are 2 and 14; the second integer that we are looking for is the square root of 100, which is 10.