If a and b are integers such that a^2− b^2 = 100, what is the greatest possible value of a?

Answer 1

arrange to a^2 = b^2 + 10^2
so a must be the hypotenuse of a right-angled triangle

look at a list of small Pythagorean triples , and look at the hypotenuse:

leg1 leg2 hypotenuse
3 4 5
5 12 13
11 60 61
19 180 181
29 420 421
any multiples of those would also work.
Consider : 3, 4, 5
we also have 6, 8, 10 ----> 10^2 = 6^2 + 8^2,
so a=10, b=8
a = 10 is the largest and only value of a which satisfies your equation.

btw, did you know you can generate Pythagorean triples if
for any m and n, m > n , and m and n have no common factor between them, then
2mn, m^2 - n^2, and m^2 + n^2 will produce a Pythagorean triple.

e.g let m= 7 and n = 3
2mn = 42
m^2 - n^2 = 40
m^2 + n^2 = 58 and 42^2 + 40^2 = 58^2

If one of m and n is even , and the other odd, you will get a UNIQUE Pythagorean triple
e.g. m = 9, n = 4
2mn =72
m^2 - n^2 = 65
m^2 + n^2 = 97 -----> 97^2 = 72^2 + 65^2

Answer 2

the greatest value of "a" corresponds with the least value of b^2
... zero is the integer with the smallest square