Asked by XXX
A number is called a perfect square if it is the square of an integer. How many pairs of perfect squares differ by 495? (Order does not matter. So, the pair "16 and 9" is the same as "9 and 16".)
Answers
Answered by
Steve
you want x^2-y^2 = 495
(x+y)(x-y) = 495
495 = 3^2*5*11
so, the factors are
1 3 5 9 11 15 33 45 55 99 165 495
so, if you have
x+y=495
x-y=1
248^2-247^2 = 495
x+y=165
x-y=3
84^2-81^2 = 495
and so on
(x+y)(x-y) = 495
495 = 3^2*5*11
so, the factors are
1 3 5 9 11 15 33 45 55 99 165 495
so, if you have
x+y=495
x-y=1
248^2-247^2 = 495
x+y=165
x-y=3
84^2-81^2 = 495
and so on
Answered by
XXX
please continue
Answered by
Steve
nope. Just take the factors in pairs, as I did, working your way in from the ends.
Actually, even that's not necessary, unless you just want to verify the results. There are 12 factors, so that means there are 6 pairs like the two I showed above.
So, there are 6 pairs of numbers whose squares differ by 495.
Actually, even that's not necessary, unless you just want to verify the results. There are 12 factors, so that means there are 6 pairs like the two I showed above.
So, there are 6 pairs of numbers whose squares differ by 495.
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