Algebra

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

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  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

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

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