The number of pairs of two digit square numbers, the sum or difference of which are also square numbers is
(1) 0
(2) 1
(3) 2
(4) 3
To do:
We have to find the number of pairs of two digit square numbers, the sum or difference of which are also square numbers.
Solution:
Two digit square numbers are $4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81$.
Let us find the difference and sum between the square numbers.
$81-64=17, 81-49=32, 81-36=45, 81-25=56, 81-16=65$
$64-49=15, 64-36=28, 64-25=39, 64-16=48$
$49-36=13, 49-25=24, 49-16=33$
$36-25=11, 36-16=20$
$25-16=9=3^2$
$16+25=41, 16+36=52, 16+49=65, 16+64=80, 16+81=97$
$25+36=61, 25+49=74, 25+64=89, 25+81=106$
$36+49=85, 36+64=100=10^2, 36+81=117$
$49+64=113, 49+81=130$
$64+81=145$
From the above data we can observe that the sum or difference of the pairs $(25, 16), (36, 64)$ form a perfect square.
There are 2 such pairs.
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