Pythagorean Palindrome

Algebra Level 4

Consider integers $a$ and $P$ that satisfy the equations

$\begin{cases} (4a+5)^2 = a^2 + [4(a+1)]^2 \\ P =(271a)^2 + [1084(a+1)]^2\\ \end{cases}$

Find sum of all value(s) of $P$ .

Image Credit: Wikimedia Pythagoras Theorem Proof

The answer is 123527762.

2 solutions

Nihar Mahajan
Mar 9, 2015

$a^2 + [4(a+1)]^2 = (4a+5)^2$

$\Rightarrow a^2 + 16a^2 + 32a + 16 = 16a^2 + 40a + 25$

$\Rightarrow a^2 - 8a - 9 = 0 \Rightarrow (a+1)(a-9) = 0 \Rightarrow a=-1 \quad or \quad a=9$

When $a=9$ we get the Pythagorean triplet $(9,40,41)$ .

When each term is multiplied by $271$ , we get $2439 , 10840 , 11111$ , which is also a Pythagorean triplet.

Comparing the above triplet and $P = (271a)^2 + [1084(a+1)]^2$ , $(a=9)$ ,

$P = (11111)^2 = 123454321$ (a palindrome).

When $a=-1$ , after substituting it in $P = (271a)^2 + [1084(a+1)]^2$ , we get

$P = (-271)^2 = 73441$

Hence , sum of all values of $P = 123454321 + 73441 =\huge\color{#D61F06}{\boxed{123527762}}$

This is a nice way to factorize the equation too.

$a^2+[4(a+1)]^2=(4a+5)^2\\ \implies a^2+[4(a+1)]^2=[1+4(a+1)]^2\\ \implies a^2+[4(a+1)]^2=1+[4(a+1)]^2+8(a+1)\\ \implies a^2-8a-9=0 \implies a=9,(-1)$

Prasun Biswas - 6 years, 3 months ago

yeah.. that's out of the normal path followed.. but really nice trick, will be helpful for large numbers.

Rishabh Tripathi - 6 years, 3 months ago

Nice! Thanks!!!!!

Nihar Mahajan - 6 years, 3 months ago

I did same!!

Dev Sharma - 5 years, 7 months ago

William Isoroku
May 29, 2015

Solve for $a$ then use substitution.

Pythagorean Palindrome

Image Credit: Wikimedia Pythagoras Theorem Proof

The answer is 123527762.

2 solutions

0 pending reports