# On combining $n$ and $n^2$ into one number

Consider the sequence $T_n$ formed by combining $n$ and $n^2$ into one number. ie. (A053061) $$T_n=\{11,24,39,416,525,636,749 \cdots\}$$ It is easy to see $$T_n= 10^{\lceil 2 \log_{10}(n) \rceil } n+ n^2$$

I looked at the sequence closely trying to find if there are any perfect squares in the sequence but wasn't able to upto $n=100$. I also was able to prove that if $n^{th}$ term is a perfect square then :

1) $n \equiv 8 \text{ or } 0 (\text{ mod } 9)$

2)In the case where $n=9m$ , m is not a square free number.

But I am unable to attack the question

Do there exist any perfect square in the sequence?

• There is something wrong with the formula you gave for $T_{n}$. As it stands, $T_{1} = 2$. On the on the hand, if by $\log n$ you are actually referring to the base $10$ logarithm of $n$ then $T_{n}$ is nothing but $n^{3}+n^{2}.$ – user2624 Jun 7 '14 at 16:01
• Not sure why people are voting to close (unless maybe it was before the post was corrected). The general sequence $B^{\lceil 2\log_B(n)\rceil}\cdot n+n^2$ seems like a perfectly reasonable sequence to study, and asking if it contains squares is also a reasonable question. And I'm pretty sure it's not going to be easy to determine, say, if the sequence contains finitely many or infinitely many squares. – Joe Silverman Jun 7 '14 at 16:50
• $a^2=b^2+10^n b$, so $a^2-(b+10^n/2)^2=10^{2n}/4$. So solutions come from factorizations of $10^n/4$. We are looking for solutions with $b$ between $10^{n-1}$ and $10^n-1$. This corresponds to factors in a certain range. – Will Sawin Jun 7 '14 at 17:44
• We can remove any extraneous factors of $10$ until $a+b$ and $a-b$ are $2\cdot5^{2n}$ and $2^{2n-3}$. Then clearly neither the sum nor the difference is in the correct range. They are much too big. So I think this is impossible. – Will Sawin Jun 7 '14 at 18:11
• @PerAlexandersson It is not about the age: planetmath.org/florentinsmarandache – Benjamin Dickman Jun 8 '14 at 0:07

There are no squares in this sequence.

Elements of this sequence are of the form $b^2+10^n b$ for $10^{n-1} \leq b^2 < 10^n$. So we are looking for a special solution to the equation $a^2 =b^2+ 10^n b$. This is an easy Diophantine equation to solve: We have $a^2 - (b+10^n/2)^2 = - 10^{2n}/4$, or

$(a+b+10^n/2)(a-b-10^n/2) = -10^{2n}/4$

The two factors on the left side have the same parity. Clearly we may assume $n>1$, so these factors both must be even. If those factors are both multiples of $20$, then $a$ and $b$ are both multiples of $10$, and we may divide $a$ and $b$ by $10$ and subtract $1$ from $n$ without changing the equation. However, it does modify the inequality. So each factor is either a power of $2$ or $2$ times a power of $5$, hence the two factors must be $2 \cdot 5^{2n}$ and $2^{2n-3}$. This gives:

$a+b+10^n/2 = 2\cdot 5^{2n}$

$a-b-10^n/2 = 2^{2n-3}$

$b= 5^{2n} - 2^{2n-4} -10^n/2$

For $n=2$, we get $575$, but $575^2>100$. For higher $n$, it's even worse: $b^2$ is much greater than $10^n$. Similarly, multiplying $a$ and $b$ by $10$ and adding $1$ to $n$ also makes $b^2$ larger, relative to $10^n$. So there are no solutions.

• can this method be extended tp prove there exist no perfect $k$ powers by combining $n$ and $n^k$ ? – Shivam Patel Jun 8 '14 at 4:15
• @ShivamPatel - I don't immediately see how, because the equation is more complicated, but there might be a way. – Will Sawin Jun 8 '14 at 12:30
• @ Will Sawin ..the can we proceed to prove that there exist no perfect squares in sequences formed by combining $n$ and $n^k$? – Shivam Patel Jun 8 '14 at 12:33