# Why every positive integer can be written as addition of 1 triangle number, 1 square number, and 1 pentagonal number

I find this interesting conjecture, maybe I'm not the first to state this conjecture. I have tested this conjecture to $100.000$ first positive integers (then my friend at Brainden forum tested it to 1 trillion), but test result is not the prove. Anybody can provide the prove or disprove this conjecture ?

the conjecture is : Every positive integer can be written as addition of 1 triangle number, 1 square number, and 1 pentagonal number

note :

Triangle numbers are generated by the formula, $T_{n} = \frac{1}{2} n(n+1)$. The first ten triangle numbers are: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, ...

Square numbers are generated by the formula, $S_{n}=n^{2}$. The first ten square numbers are: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81,...

Pentagonal numbers are generated by the formula, $P_{n}=\frac{1}{2}n(3n-1)$. The first ten pentagonal numbers are: 0, 1, 5, 12, 22, 35, 51, 70, 92, 117,...

10 first positive integer which follow the conjecture :

Numbers = Triangle + Square + Pentagon

1 = 0 + 0 + 1

2 = 0 + 1 + 1

3 = 1 + 1 + 1

4 = 0 + 4 + 0

5 = 0 + 0 + 5

6 = 0 + 1 + 5

7 = 1 + 1 + 5

8 = 3 + 0 + 5

9 = 0 + 4 + 5

10 = 0 + 9 + 1

• This is very interesting. I know absolutely nothing about number theory, but questions like this make me really wish I did. Aug 11, 2016 at 2:47
• This is a special case of a more General hypothesis. Curves for triangular numbers - there are always solutions math.stackexchange.com/questions/794510/… Aug 11, 2016 at 4:17
• suppose we have 3 different number, We can have 6 different triangle + square + pentagonal numbers, from combination of the numbers. But we can only have 1 square + square + square number. I think that's why some integers can't be written from three squares. Aug 13, 2016 at 3:13

Friday: found a new arXiv item near to this, Haensch Kane. There has been a fair amount of activity in this general area of late, for example Kane and Sun on mixed sums of triangular numbers and squares. Apparently Guy showed that every positive number is the sum of three pentagonal numbers.

Apparently this is discussed in SUN. My first impression is that he could not prove it either; he did prove it if all the variables are allowed to be negative when needed.

Thursday: below is a proof that the ternary form $x^2 + 6 y^2 + 12 z^2$ does something desirable. The original question follows (maybe) from this. It will take me a bit to type that last part in.

Alrighty, using $t$ for triangle, $s$ for square, $p$ for pentagon, we want $$\frac{t^2 + t}{2} + s^2 + \frac{3p^2 - p}{2 } = n.$$

Multiply by $8$ and add $1$ to both sides, we have $$(2t+1)^2 + 8 s^2 + 12 p^2 - 4p = 8n+1$$

Multiply by $3$ and add $1$ to both sides, we have $$3(2t+1)^2 + 24 s^2 + (6p-1)^2 = 24n+4.$$

One fourth of $24n+4$ is $6n+1.$ Take $$u^2 + 12 v^2 + 6 w^2 = 6n+1,$$ $$u^2 + 3 (2 v)^2 + 6 w^2 = 6n+1.$$

Hmmmm.

Let me put what I've got next, which i assumed was enough last night: Take $u,v,w \geq 0.$ Note $u$ is odd and prime to $3.$

If $u \equiv 2 \pmod 3,$ $$(u+6v)^2 + 3 (|u-2v|)^2 + 24 w^2 = 24 n+4.$$

If $u \equiv 1 \pmod 3$ and $u < 6v,$ $$(6v -u)^2 + 3 (u+2v)^2 + 24 w^2 = 24 n+4.$$

If $u \equiv 1 \pmod 3$ and $u > 6v,$ more work is needed. Hmmm.

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Wednesday: Make an answer; this is my area. Your question follows from the fact that, given integer $n \geq 0,$ we can always write $$6n+1 = u^2 + 6 v^2 + 12 w^2$$ in integers.

The ternary form $u^2 + 6 v^2 + 12 w^2$ is not regular. Up to a million, there are some ten numbers that are not represented by the form, but instead are represented by the other form in the genus, that being $3 u^2 + 4 v^2 + 6 w^2.$ These sporadic numbers are, however, all multiples of three.

  3         39         51         69        165        219
399        561        771       1941


Taken together, the forms of the genus represent all positive integers except $3k+2$ and $4^k (16m+14).$ Compare the form $1,3,6$ in Dickson's table:

The composition of the genus is shown in this excerpt

B.-I.discr = -288 = -1 *2^5 *3^2
1^-1 3^-2
[2^1 4^1 8^1]_3
616: 1 6 12 0 0 0
631: 3 4 6 0 0 0
2^-2_4 16^-1_3
613: 1 3 24 0 0 0
637: 4 4 7 2 4 4
2^-2_2 16^-1_5
632: 3 4 7 4 0 0
2^2_2 16^1_1
619: 1 9 9 6 0 0


Proof of fact above....

Lemma (Burton Wadsworth Jones, 1928 dissertation) If we have a nonzero integer $k$ with $k=r^2 + 2 s^2,$ and $k$ is a multiple of $3,$ there is another representation $k = x^2 + 2 y^2$ with both $x,y$ prime to $3.$

We can also prove this by induction on the power of $3$ dividing $k.$

Theorem: if $m \equiv 1 \pmod 6$ and $n > 0,$ we can always write $m = u^2 + 6 v^2 + 12 w^2$ in integers.

Proof We may assume that $m$ is represented by the other form of the genus, so we have $m = 3 u^2 + 4 v^2 + 6 w^2.$ It follows that $v \neq 0 \pmod 3.$ We have four similar formulas,

$$9m = (3u+6w)^2 + 6(-u + 2v+w)^2 + 12 (-u-v+w)^2$$

$$9m = (3u-6w)^2 + 6(-u + 2v-w)^2 + 12 (-u-v-w)^2$$

$$9m = (3u+6w)^2 + 6(u + 2v-w)^2 + 12 (u-v-w)^2$$

$$9m = (3u+6w)^2 + 6(u + 2v+w)^2 + 12 (u-v+w)^2$$

If at least one of $u,w$ is not divisible by $3,$ then at least one of the four formulas has all three items that are being squares divisible by $3.$ Therefroe we may divide them by $3$ and move $9m$ to $m$ itself.

If both $u,w$ are divisible by $3,$ then the Jones Lemma says that we can revise the representation so that both $u,w$ are prime to $3.$ as a result, in the revision, at least one of the four formulas is made up of numbers divisible by $3.$

$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$

• Thank you. I was thinking to approach this with much more elementary techniques, which I think you statement "is not regular" says will not work. Aug 11, 2016 at 4:29
• @RossMillikan I have put in the proof for the conjecture. Some of it needs to be taken for granted, the composition of this particular genus, the numbers represented by the entire genus, and the fact that any eligible number is represented by one of the forms of the genus. Aug 11, 2016 at 4:56
• What is the significance of the row of circles at the bottom? Aug 11, 2016 at 17:59
• @BrianTung it means that proof is ended. Like Q.E.D. Aug 11, 2016 at 18:02