What number when squared and either increased or decreased by 5 gives a number that is a perfect square?
The number we're looking for, let's call it [itex]a[/itex]. Let's call the perfect square [itex]b^2[/itex]. So:scox said:What number when squared and either increased or decreased by 5 gives a number that is a perfect square?
Gokul43201 said:Since integral squares are separated by odd numbers or multiples of 4(add multiples of 8 to convert from necessary to sufficient), the only way to get a difference of say, 10 (see abertram's frustration) is to find a functioning difference that is a multiple of 10 (such as 40, between 9 and 49) - we know that this exists, since 40 is a multiple of 8 - and divide by 4.
Hence, 49/4 and 9/4 are the outer squares and the solution is the square root of the mean, ie : sqrt(29)/2
scox said:Well thanks for all the help everyone. When I get the answer at the end of the semester I will share it with all of you, it has stumped me and at the same time made me a little mad....lol. According to the teacher my answer is correct, but I dont know how concidering (29)/2 is not a perfect square. Again thanks alot everyone!!
I am anxious to figgure this out.