Medium numbers.

Number Theory Level 3

How many triples $(a, b, c)$ of integers exists such that

$\large\ a^4 + b^3 = c^2 ?$

2 0 1 None of the others Infinitely many 2019

5 solutions

Zee Ell
Jun 6, 2019

It is easy to find the smallest "all-positive" triplet:

$1^4+2^3=3^2$

Now, it is easy to show, that there are infinitely many triplets, by multiplying our equation by the 12th power ( LCM(2, 3, 4) = 12) of the positive integer k:

$k^{12} (1^4+2^3) = 3^2 × k^{12}$

$(k^3)^4 + (2k^4)^3 = (3k^6)^2$

Now, we can see, that we have infinitely many triplets in the form of:

$(k^3, 2k^4, 3k^6), k \in \mathbb {N}$

Hence, our answer should be:

$\boxed { \text { Infinitely many } }$

Remark:

There are other base triplets (where GCD(a, b, c) = 1 (a.k.a. coprimes)) in existence. Some of them can be found in Chris Lewis's solution. However, just one of them (e.g. the (1, 2, 3) we used here) is sufficient for the proof that there are infinitely many such triplets.

Cantdo Math
Apr 15, 2020

Just take b=0 and c to be a perfect fourth power.

Pablo Sanchez
Jul 9, 2019

I wrote it as c^2 - a^4 = b (b^2) So: (c-a^2)(c+a^2)=b (b^2)

c-a^2=b

c+a^2=b^2

From that you get that

b=[ 1+sqr(1+8c)]/2

And it takes little effort to figure out that

C = [n(n+1)]/2

For every n.

So with the values of c and b you can figure out a from the first equations. From that is very obvious that there are infinite solutions.

After a little time solutions can be in the form

a=sqr((n(n+1))/2)

b=n

c= (n(n+1))/2

For any n bigger than 1

Pablo Sanchez - 1 year, 11 months ago

Edwin Gray
Jun 19, 2019

Let b = 0, c = a^2

Chris Lewis
Jun 6, 2019

Infinitely many triples of the form $(a,b,c)=(t,0,t^2)$ work.

Are there infinitely many solutions excluding zero?

I found the same thing, a^4+0^3 = c^2
If we add the word 'positive' before integer, I believe we reduce it to a finite amount. I found 15 triplets when A and B are both less than or equal to 100. Maybe we can investigate these further and see if theres some recurrence.
1^4+2^3 = 3^2
5^4+6^3 = 29^2
5^4+75^3 = 650^2
6^4+9^3 = 45^2
6^4+72^3 = 612^2
7^4+15^3 = 76^2
8^4+32^3 = 192^2
9^4+27^3 = 162^2
9^4+54^3 = 405^2
15^4+100^3 = 1025^2
17^4+42^3 = 397^2
33^4+27^3 = 1098^2
35^4+50^3 = 1275^2
40^4+96^3 = 1856^2
44^4+80^3 = 2064^2

Kyle T - 2 years ago

I think you should change the word integers to positive integers, otherwise the problem becomes too easy. (Well the answer remains same on both cases, but allowing zero gives some trivial solutions such as a=t, b=0, c=t^2 as many people found out.)

Arunsoumya Basu - 1 year, 11 months ago

Medium numbers.

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