Math Is Fun Forum

Prakash Panneer

Member

Registered: 2006-06-01

Posts: 110

Integers

The positive integers a, b are such that 15a + 16b and 16a - 15b areboth squares of positive integers. What is the least possible value thatcan be taken on by the smaller of these two squares?

Thanks in advance

---

Letter, number, arts and science
of living kinds, both are the eyes.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Integers

I've worked out that for the smaller square to be 1, then the larger square has to be of the form 481n + 31, where n is a positive integer. I've used Excel to work out all of those for n up to 20000, and none of them are squares. And yet I can't see why it's impossible for 481n+31 to be a square. It seems like if we go high up enough then we'll eventually find one. Can anyone make a proof that we won't?

Edit: More generally, for the smaller square to be k, then the larger square has to be of the form 481n+31k. Not sure if that helps at all.

---

Why did the vector cross the road?
It wanted to be normal.

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Integers

Very interesting question!Pity that I can seldom be online now , but  I have thought about it for hours , This is what I get :
15a + 16b =m^2   16a - 15b =n^2

a=14911 b=481 n=m=481

I don't know if it's right

after substitution of b ,  I get  a=(15m^2+16n^2)/481
then I thought a can only be an integer when m^2=481^2=n^2 .so.....lol~

---

Numbers are the essence of the Universe

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers

Hi all;

Old but good:

Can anyone make a proof that we won't?

If all the squares have to be of the form 481n + 31 then it is easy to prove that there are no such squares.

So there is no square that is of the form 481n + 31 when n is an integer.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.