Integer squares and their doubles

I noticed one day that 5 squared is 25 which is almost exactly half of 7 squared which is 49. Then I noticed that 7 squared is 49 which is almost exactly half of 100 which is 10 squared. I fount this mildly interesting, if not significant, so I looked at some other squares that are almost exactly half of other squares. Here are some examples:

2 * 2^2 = 2 * 4 = 8, 3^2 = 9 2 * 5^2 = 2 * 25 = 50, 7^2 = 49 2 * 12^2 = 2 * 144 = 288, 17^2 = 289 2 * 29^2 = 2 * 841 = 1682, 41^2 = 1681 2 * 70^2 = 2 * 4900 = 9800, 99^2 = 9801 2 * 169^2 = 2 * 28561 = 57122, 239^2 = 57121 2 * 408^2 = 2 * 166464 = 332928, 577^2 = 332929 2 * 985^2 = 2 * 970225 = 1940450, 1393^2 = 1940449 2 * 2378^2 = 2 * 5654884 = 11309768, 3363^2 = 11309769 2 * 5741^2 = 2 * 32959081 = 65918162, 8119^2 = 65918161

Strangely enough, I haven't found any cases where twice one square is exactly another square, just almost. Are there any such numbers? If so, what are they? If not, why not?

Bazerko Bob

No. Suppose there was at least one such pair. Then there must be a smallest pair; call them A^2 and B^2.

B^2 = 2*A^2 2*A^2 is even B^2 is even B is even (if it were odd, then B^2 would also be odd) B = 2*N for some integer N (2*N)^2 = 2*A^2 4*N^2 = 2*A^2 2*N^2 = A^2

But now (N^2, A^2) is a smaller pair than (A^2, B^2). This is a contradiction. Therefore, there is no such pair.

A similar argument proves that the square root of 2 is rational. If it were rational, then it could be expressed as the ratio of two integers; the smallest such positive integers are B and A, and the rest of the proof follows the pattern shown above.

Think about the following: sqrt(2) = 1.4142135623731 3 / 2 = 1.5 7 / 5 = 1.4 17 / 12 = 1.416667 41 / 29 = 1.413793 99 / 70 = 1.414286 239 / 169 = 1.414201 577 / 408 = 1.414216 1393 / 985 = 1.414213 3363 / 2378 = 1.414214 8119 / 5741 = 1.414214

There aren't any:

Suppose a and b are the smallest integers such that: a*a = 2 * b * b a must be even, so a= 2c so 2c * 2c = 2*b*b so 2* c*c = b*b Thus b and c satisfy the same condition, which contradicts the premise that a and b are the smallest such integers.

So there is no solution (This is essentially the same as the proof that sqrt(2) is irrational)

This is, of course, basically correct. But it contains an error. Detect this error: and find a pair of squares of integers, one of them double the other.

You're right, of course, technically. After all, the title of the thread refers to _integer_ squares. And it had not been stated, although I suspect that most had assumed, that we were concerned only with _positive_ integers, for which the argument is correct.

0^2 = 2 * 0^2

As others have posted, of course not, because sqrt(2) is irrational. Anyway, your almost-squares are a simple case of a Pellian equation. If the pair (a,b) satisfies 2*a^2 = b^2 +- 1, then so does the pair (a+b, 2a+b). Your examples can be found by iterating this rule, starting from the pair (0,1).

Bob Morris originally asked for 'cases where twice one square is exactly another square', implying that A^2 != B^2. This is not true of A=0, B=0.

It's interesting that I had to read 'cases where twice one square is exactly another square' _several_ times before I was able to understand how you thought it could imply that A^2 != B^2. The crux of the matter is 'another'.

The proof I had in mind was b^2/a^2=2 so b/a=sqrt(2) for which no integers have been found, but I like Ed's, too. (I don't remember the proof that sqrt(2) is irrational, but I rely on it.)

Bazerko Bob

The proof that sqrt(2) is irrational is virtually identical, and starts with the line: 'Suppose that sqrt(2) is rational, and equal to the fraction a/b, where a,b are relatively prime integers'