sqrt(a) + sqrt(b) = sqrt(x)

How many possible values of x less than or equal to 1000 are there?

After 4 pages of scribbled calculations, here is what I came up with:

The equation we were given is sqrt(a)+sqrt(b)=sqrt(x). As noted previously by Charlie, this means that a+b+2sqrt(ab)=x, and so x-(a+b)=2sqrt(ab), so therefore the product ab must be a perfect square. So (with this in mind) for every possible value of a, I considered all corresponding possible values for b, and worked out what values x would have to take in those cases.

If a=1, then b must be a perfect square. So

b=1, 4, 9, 16, 25, 36, ... and these cases give us

x=4, 9, 16, 25, 36, 49, ... (check these results)

We have a pattern! In general, a=1, b=n^2, and x=(n+1)^2.

Now, if a=2, then b must be twice a perfect square. So

b=2, 8, 18, 32, 50, 72, ... and these yield

x=8, 18, 32, 50, 72, 98, ...

Here, a=2, b=2*n^2, and x=2(n+1)^2.

Next, if a=3, then b must be three times a square. So

b=3, 12, 27, 48, 75, 108, ... and these give us

x=12, 27, 48, 75, 108, 147, ...

Notice what is happening here! If a=3, for example, b will then have to take the form 3*n^2 (for some positive integer n), and so we have to have

sqrt(3)+sqrt(3*n^2)=sqrt(x)

sqrt(3)+n*sqrt(3)=sqrt(x)

(n+1)sqrt(3)=sqrt(x)

sqrt(3(n+1)^2)=sqrt(x)

so x=3(n+1)^2

This pattern works for any original choice of a. So the possible values for x will be (n+1)^2, 2(n+1)^2, 3(n+1)^2, ... etc. In fact all positive multiples of all perfect squares greater than 1 will work. (Of course, let's not forget that we are only considering values of x less than or equal to 1000). That is, x can be

4, 8, 12, 16, 20, ...

9, 18, 27, 36, 45, ...

16, 32, 48, 64, 80, ...

25, 50, 75, 100, 125, ...

36, 72, 108, 144, 180, ... etc.

So, how many possible values for x are there less than 1000? This is a problem in combinatorics. There are 250 multiples of 4 that are not greater than 1000. Also, there are 111 multiples of 9 (but we have already counted floor(111/4)=27 of them, which means this only adds 111-27=84 more possibilities.) We don't have to consider multiples of 16 as they have already been counted as multiples of 4. There are 40 multiples of 25, but we have already counted 10 of them as multiples of 4, and 4 of them as multiples of 9... but let's keep the one which is a multiple of 4*9=36. So this adds 40-10-4+1=27 more cases.

Continue in this way, all the way up to the single multiple of 31^2=961, which is the last possible case. Adding up the possibilities gives me 250+(111-27)+(40-10-4+1)+(20-5-2)+(8-2)+(5-1)+3+2+1+1+1=392 possible values for x. (Notice that I only had to consider multiples of squares of prime numbers here; the others had been counted before!)

Of course, now comes the worrying point. My answer disagrees with Charlie's!!! I agreed completely with everything that he wrote in his posts, but I don't want to try and comprehend his computer program... Can anyone see a mistake in my argument here? I am very interested to know what the correct solution is!

A wonderful problem!! Thanks Jer (coincidentally, my initials are Jer!)