find the smallest positive integer $a,$ greater than 1000, such that the equation
\sqrt a - \sart a-x has a rational root.

84

Step-by-step explanation:

So the number of divisors is:

σ=(1+na1)(1+na2)...(1+nak)

Suppose that the target number n has 4 distinct prime factors. The smallest such number is n=2×3×5×7=210. The number of divisors of nn is (1+210)4 which is obviously greater than 106. There is no need to consider any other number n with 4 or more disitnct prime factors and we already know that the solution is not greater than 210.

Ok, let us consider numbers with 3 distinct prime factors. Start with the smallest ones: 30, 42, 60, 66, 70, 78, 84... The checks are quick and simple, you won't waste too muct time (thanks EspeciallyLime for providing the right sequence, I have skipped a few possible values initially).

Just a few examples:

n=2×3×5=30, number of divisors is 313<106, solution discarded.

…

n=22×3×5=60, number of divisors is 121×61×61<106, solution discarded.

…

n=22×3×7=84, number of divisors is 169×85×85=1221025, stop!

There is no need to consider any other number n with three distinct prime factors because they are all bigger than 84.

Let's move to numbers with two distinct prime factors. Obviously, we'll have to check only numbers less than 84. Such numbers will have the following form:

n=pa11pa22

nn=pna11pna22

...with the number of divisors equal to:

σ=(1+na1)(1+na2)>1000000(1)

So at least one of the factors will have to be greater than 1000000−−−−−−−√=1000. But:

1+nai>1000⟹ai>999n>99984>11

But this is nonsense, because the smallest number with two prime factors and ai=12 is 212×3 which is way above our best value so far (84).

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: