Suppose that only one of the folowing pairs (x,y) yields the positive integer sqr root(x^2 + y^2). Then..?

Question

(A) x=25530, y=29464 (B) x=37615, y=26855 (C) x=15123, y=32477; (D) x=28326, y =28614 (E) x=22536, y=27462

(Note: No calculator is allowed for this question)

Tks a lot

Answer 1

No calculator. x^2 + y^2 must be a perfect square.

The ones digit of the square of a number is determined by the ones digit of the number. (Below on both sides is "number ending in.")
0^2 -> 0
1^2 -> 1
2^2 -> 4
3^2 -> 9
4^2 -> 6
5^2 -> 5
6^2 -> 6
7^2 -> 9
8^2 -> 4
9^2 -> 1
A perfect square of an integer ends in 0,1,4,5,6, or 9.

Then, looking at the ones digits of your options above:

0^2 + 4^2 -> 0 + 6 -> 6 still possible;
5^2 + 5^2 -> 5 + 5 -> 0 still possible;
3^2 + 7^2 -> 9 + 9 -> 8 not possible;
6^2 + 4^2 -> 6 + 6 -> 2 not possible;
6^2 + 2^2 -> 6 + 4 -> 0 still possible.

We have eliminated two answers as possibilities.

Now,
If a perfect square of a number ends in 0, then 2 divides the square, and thus 2 divides the number (since 2 is prime). Thus 2^2 = 4 divides the square. Similarly, 5, and thus 5^2 = 25 divides the square. Thus a perfect square ending in "0" is divisible by 2^2 * 5^2 = 100, and thus must end in *two* zeros.
(Or, note that only 0^2 -> 0, and thus the original number must be divisible by 10, and hence the square is divisible by 100.)

The last two digits are determined by the last two digits only. Then for (B) we have

15^2 + 55^2 -> 25 + 25 -> 50 does not end in "00," and
36^2 + 62^2 -> 96 + 44 -> 40 does not end in "00"

eliminates (E).

Thus the only answer left, (A), is your answer.

(A calculator confirms it. 25530^2 + 29464^2 = 38986^2.)

Hope this helps.

Answer 2

C and D are eliminated because the last digit of the square would be 8 and 2 respectively. That can't happen.

More generally, the last two digits of the square are the same as those of the square of the number formed by the last two digits..

So in B x^2 ends the same way as 15^2 = 225, namely 25, while 55^2 = 5*11*5*11 = 121*25 also ends with a 25. But no perfect square ends 25+25 = 50. So B is out. Similarly in E you'll find that the last two digits of x^2 + y^2 are 60.

So the only possibility is A.