# Are there infinitely many palindromes whose sum is a perfect square?

I am just curious on wether there are infinitely many palindromes say $p_1$ and $p_2$ satisfying:

$p_1^2+p_2^2$ is a perfect square with $\gcd(p_1,p_2)=1$.

I believe that there are some but, are there infinitely many of them?

• So far, I only found the pairs $(3/4)$ and $(464/777)$ Feb 15, 2015 at 10:14
• Thanks peter, I suspect that there are only finite of them, but a general proof still awaits. Feb 15, 2015 at 10:17
• I actually want to prove my claim, but I dont know how to. Feb 15, 2015 at 10:20
• Does the pair $(0/1)$ also count ? Feb 15, 2015 at 10:24
• No peter, we start at 1. Feb 15, 2015 at 10:25

Yes, there are infinitely many. For any $n\in\mathbb{N}$ let \begin{align} a_n&=3+4\times10^{n}+7\times10^{2n}+4\times10^{3n}+3\times10^{4n}\\ b_n&=4+2\times10^{n}+8\times10^{2n}+2\times10^{3n}+4\times10^{4n}\\ c_n&=5+4\times10^{n}+11\times10^{2n}+4\times10^{3n}+5\times10^{4n} \end{align} Then $a_n$ and $b_n$ are palindromes and $$a_n^2+b_n^2=c_n^2.$$ Moreover, $\gcd(a_n,b_n)=1$. Let \begin{align} A_n&=7+8\times10^{n}+8\times10^{2n}+8\times10^{3n}\\ B_n&=-\frac12\bigl(10+21\times10^{n}+22\times10^{2n}+12\times10^{3n}\bigr) \end{align} Then $$A_n\,a_n+B_n\,b_n=1.$$

I found this identity doing a brute force search. Any palindrome with an even number of digits is divisible by $11$, so one (or both) of the $p_i$ must have an odd number of digits. I have searched for $p_1$ with $2\,k+1$ digits, $1\le j\le 7$. These are the results. Included are also some examples of $p_1$ with $17$ digits. $$\begin{array}{ll} 313 & 48984 \\ 464 & 777 \\ 25652 & 55755 \\ 34743 & 42824 \\ 52625 & 80808 \\ 80308 & 5578755 \\ 2152512 & 575575 \\ 2532352 & 5853585 \\ 5679765 & 23711732 \\ 304070403 & 402080204 \\ 341484143 & 420282024 \\ 345696543 & 422282224 \\ 355949553 & 690019910096 \\ 359575953 & 401141104 \\ 27280108272 & 55873637855 \\ 3004007004003 & 4002008002004 \\ 3044529254403 & 4022208022204 \\ 3410048400143 & 4200028200024 \\ 3414249424143 & 4202028202024 \\ 3450569650543 & 4224448444224 \\ 6381414141836 & 778233332877 \\ 395734505437593 & 426982282289624 \\ 404990565099404 & 747709181907747 \\ 461781161187164 & 778676101676877 \\ 30004000700040003 & 40002000800020004 \\ 30040410801404003 & 40020200800202004 \\ 30044412921444003 & 40022200800222004 \\ 30081842624818003 & 40041401210414004 \\ 30401040804010403 & 40200020802000204 \\ 30405060906050403 & 40202020802020204 \\ 32682698889628623 & 46020004840002064 \\ 34100004840000143 & 42000002820000024 \\ 34104204940240143 & 42002002820020024 \\ 34140434943404143 & 42024404840442024 \\ 34505056965050543 & 42244646864644224 \\ \end{array}$$

• Thanks a lot Prof., if it is okay with you sir, can we have a detailed discussion about this topic? Thanks. Feb 23, 2015 at 15:20
Following Lucian's hint, I found a list of them at http://www.worldofnumbers.com/pythago.htm but many of the ones listed there fail the relative primality test. But not all: $313^2+48984^2=48985^2$, $34743^2+42824^2=55145^2$, probably a few others that I didn't test for relative primality.
• Its because of scarcity of numbers that satisfies this as an integer $n$ goes large sir. Lacking of computer softwares is a hindrance about my conjecture sir. Feb 15, 2015 at 11:53
• FYI: Fifteen of the thirty-nine triples on the World of Numbers page are relatively prime: $(3, 4, 5)$, $(464, 777, 905)$, $(313, 48984, 48985)$, $(34743, 42824, 55145)$, $(25652, 55755, 61373)$, $(52625, 80808, 96433)$, $(575575, 2152512, 2228137)$, $(80308, 5578755, 5579333)$, $(2532352, 5853585, 6377873)$, $(5679765, 23711732, 24382493)$, $(304070403, 402080204, 504110405)$, $(341484143, 420282024, 541524145)$, $(345696543, 422282224, 545736545)$, $(359575953, 401141104, 538710545)$, $(27280108272, 55873637855, 62177710753)$