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X^2+X^3= X^2(X+1)
so, x^2 perfect sq. so, x+1 also should be perfect sq.
so x= 3,8,15,......,80,99

i think this is d ans.
correct me if i m wrong.
hope this helped :)

1 5 1
:D sorry about that. thanks for d ans :)
bt u wrote x can take values starting from 1 na?
u gave d range 1-100
oh yes.... your are crrecr
The Brainliest Answer!
X*x + x*x*x = y^2
So, x^2 + x^3 = y^2
So, x^2(x +1) = y^2
So, x* root(x+1) = y [Sqroot both sides]

For y to be an integer, root(x+1) has to be an integer. So x is a predecessor of a perfect square.

There you go...

To answer the question, there would be 9 such integers, namely 3, 8, 15, 24, 35, 48, 63, 80, 99...
1 5 1
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