Are there infinitely-many numbers that are both square and triangular?

Question

I just started to read "Friendly Introduction to Number Theory" but I am getting stuck in the first exercises.

1.1. The first two numbers that are both squares and triangles are 1 and 36. Find the next one and, if possible, the one after that. Can you figure out an efficient way to find triangular–square numbers? Do you think that there are infinitely many?

https://www.math.brown.edu/~jhs/frintch1ch6.pdf

I found how to find out the number which is both square and triangle. (don't know if this is effective way)

https://github.com/y-zono/friendly-introduction-number-theory/blob/master/01/1-1/main.go

However how can I answer "Do you think that there are infinitely many?"? I think I need to find the formula but no idea yet. Can you please help me?

Answer 1

Have a look at https://en.wikipedia.org/wiki/Square_triangular_number So, the answere is yes!

Answer 2

The formula for triangle numbers is on page 9. Create an equation with the left being that formula using x, and the right the formula for squares, using y. X and y are not equal, but you can rearrange terms. Ask, when is the left divisible by something on the right? Are we sure that such divisibility exists for higher numbers?

Better, assume we have such an x and y. Can we always add something to each to get the next number?

Answer 3

Of course the solution of equation:

$$Y^2=\frac{X(X\pm1)}{2}$$

Defined solutions of Pell's equation: $$p^2-2s^2=\pm1$$

But it is necessary to write the formula describing their solutions through solving Pell's equation:

$$X=p^2+4ps+4s^2$$

$$Y=p^2+3ps+2s^2$$

And more.

$$X=2s^2$$

$$Y=ps$$

$p,s$ - These numbers can be any character. If you need to have a solution of the equation: $$Y^2=\frac{X(X\pm{a})}{2}$$

It is necessary to substitute into the formulas uravneniyaPellya solutions: $$p^2-2s^2=\pm{a}$$

Answer 4

There is a closely related problem. Do their exists an infinite number of triples (x,y,z) such that x,y,and z are consecutive integers each being expressible as the sum of two not necessarily distinct integer squares? For example $$0=0^2+0^2;1=0^2+1^2;2=1^2+1^2$$ $$8=2^2+2^2; 9=0^2+3^2; 10=3^2+1^2$$.

In the latter case, the triple takes the form $2m^2+2=n^2+1$, or $$n^2-2m^2=1$$ Pell's equation again.

So if you can construct or prove the existence of an infinite number of such triples, you have found an infinite number of square triangular numbers.

Answer 5

Assume that $$\frac{n(n+1)}{2}=m^2\iff n(n+1)=2m^2$$ Since $\gcd(n,n+1)=1$ we need to factor $m^2$ as $m^2=a^2b^2$ with $\gcd(a,b)=1$. Assume without loss of generality that $a^2\leq b^2$, then there are 3 cases to consider given that now $$n(n+1)=2a^2b^2$$

1. $n=a^2, n+1=2b^2$. Then $$1=2b^2-a^2\geq b^2$$ so $a=b=\pm 1$.

2. $n=2a^2, n+1=b^2$. Then $$1=b^2-2a^2$$ One solution to this is $(a,b)=(2,3)$, and we can always generate a new solution by $$(a',b')=(2ab, b^2+2a^2)$$ The motivation for this is to consider the equation $1=b^2-2a^2$ over the ring $\Bbb Z[\sqrt 2]$, with norm $N(a+b\sqrt 2)=a^2-2b^2$. Then $N(\alpha\beta)=N(\alpha)N(\beta)$, in particular if $N(\alpha)=1$ then $N(\alpha^n)=1$.

3. $n=b^2, n+1=2a^2$. Then $$1=2a^2-b^2$$ multiply by $2$ to make the equation into $$2=(2a)^2-2b^2=(2a+\sqrt 2 b)(2a-\sqrt 2 b)\in\Bbb Z[\sqrt 2]$$ one solution to this equation is $(a,b)=(1,1)$. To generate more solutions we may multiply by any unit $u$, for instance $u=3+2\sqrt 2$. $(2a+\sqrt 2 b)(3+2\sqrt 2)=6a+4b+(4a+3b)\sqrt 2$. $2=(10+7\sqrt 2)(10-7\sqrt 2)$, so $$1=2\cdot 5^2-7^2$$ Therefore $$(a',b')=(3a+2b, 4a+3b)$$ and so there are also infinitely many solutions in this case also.