Question 197753
Call the larger number {{{a}}}
Call the smaller number {{{b}}}
given:
{{{64*(a - b) = a^2 - b^2}}}
{{{64*(a - b) = (a + b)(a - b)}}}
{{{64 = a + b}}}
I now know that the sum of the numbers is {{{64}}}
I can think of the right side of the equation as
(hypotenuse)^2 - (one leg)^2 of a right triangle
where {{{a}}} is the hypotenuse and {{{b}}} is the leg
The left side of the equation must be {{{c^2}}}
where {{{c = sqrt(64*(a - b))}}}
From above, {{{b = 64 - a }}}, so
{{{c = 8*sqrt(a - (64 - a))}}}
{{{c = 8*sqrt(2a - 64)}}}
{{{c}}} must be a whole number, since the right side
of the original equation is a whole number
That means that {{{sqrt(2a - 64)}}} must be a whole number
Note that if {{{a = 32}}}, then {{{c = 0}}} and
{{{c^2 = 0}}}, {{{a^2 = b^2}}}, {{{a = b}}} and that is
not allowed by the problem.
So {{{a > 32}}}
I'll try {{{a = 33}}}
{{{b = 64 - a}}}
{{{b = 64 - 33}}}
{{{b = 31}}}
{{{c = 8*sqrt(66 - 64)}}}
{{{c = 8*sqrt(2)}}}
Not allowed, since {{{c}}} has to be a whole number
I'll try {{{a = 34}}}
{{{b = 30}}}
{{{c = 8*sqrt(68 - 64)}}}
{{{c = 8*sqrt(4)}}}
{{{c = 16}}}
{{{c^2 = 256}}}
{{{a^2 - b^2 = 34^2 - 30^2}}}
{{{a^2 - b^2 = 1156 - 900}}}
{{{a^2 - b^2 = 256}}}
So, {{{c^2 = a^2 - b^2}}}
{{{64*(a - b) = a^2 - b^2}}}
{{{64*(34 - 30) = 256}}}
{{{64*4 = 256}}}
{{{256 = 256}}}
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{{{a + b = 64}}}
The teacher had 64 old math books