Perfect square?

Question

How does one find the lowest k such that k^2+14k is a perfect square, without guess and check/

NOTE

I got it, if we let that expression equal m^2, then 196+4m^2 is a perfect square, so 49+m^2 is a perfect square, we need to find m then. so we're looking for the smallest pythagorean triple containing 7.. 7, 24, 25, so m = 24. then k=18.

okay, nevermind.

so my new question is would number theory be faster?

Answer 1

let us have k^2 + 14k = m^2 and let m = n+7

so k^2 + 14k = n^2 + 14n + 49

or (k^2-n^2) + 14 (k-n) = 49

or, (k-n) (k+n+14) = 49

but since we are dealing with integers, we must have the LHS as factors of 49; ie. 1 * 49 or 7*7

so if k-n=1 and k+n+14=49 then k=18, n = 17 (solution 1)
if k-n=49 and k+n+14=1 then k = 18 and n= - 31 (solution 2)
if k-n=7 and k+n+14 =7 then k=0 and n=-7 (solution 3)

corresponding values of m (m=n+7)

k=18 and m=24
k=18 and m= -24 ( - ve sign wont matter)
k=0 and m=0

so the least value of k is obviously 0; non-zero solution is k=18 when k^2+14k equals 24^2; and that is the only solution

Answer 2

find half of 14, square it, and add it to the equation

k^2 + 14k + 49

whereas k can be any value.

The method i used is also applied in completing the square.