Prove that $3n6n3^2+4n8n4^2$ yields a perfect square.

Question

Let $n\geq 0$ be a nonnegative integer.

I observed that the expression $3n6n3^2+4n8n4^2$ is always a perfect square where $n$ represents number of "0". And I had verified it by using a calculator for at at most 7 digit composition of $3n6n3$ and $4n8n4$. That is:

$n=0:363^2+484^2=605^2$

$n=1:30603^2+40804^2=51005^2$

$n=2:3006003^2+4008004^2=5010005^2$

This computations put me on the affirmative side that in general as $n$ runs through positive integers the given sum always yields a perfect square.

But when im trying to prove it, I dont know where I will start.

Can someone help me prove that the expression $3n6n3^2+4n8n4^2$ always yields a perfect square? Thanks a lot.

Answer 1

In more usual notation, your observation is that whenever $b$ is a power of $10$, the number $$ (3b^2+6b+3)^2 + (4b^2 + 8b + 4)^2 $$ is always a perfect square.

This holds even if $b$ is not a power of $10$. We have $$ \begin{align} &(3b^2+6b+3)^2 + (4b^2 + 8b + 4)^2 \\ ={}& (3(b^2+2b+1))^2 + (4(b^2+2b+1))^2 \\ ={}& 9\cdot(b^2+2b+1)^2 + 16 \cdot (b^2+2b+1)^2 \\ ={}& (9+16)\cdot (b^2+2b+1)^2 \\ ={}& 25\cdot (b^2+2b+1)^2 \\ ={}& (5(b^2+2b+1))^2 \end{align} $$

It also happens that $b^2+2b+1$ itself is a perfect square (namely of $b+1$), but that is not necessary for the rewriting to work.

Answer 2

$$ 11^2 = 121 $$ $$ 101^2 = 10201 $$ $$ 1001^2 = 1002001 $$ $$ 10001^2 = 100020001 $$ and $$ 3^2 + 4^2 = 5^2 $$

Answer 3

If you insert $n$ zeroes in the number $11$, this number is expressible as

$$N(n) = 10^{n+1}+1.$$

This number squared is

$$[N(n)]^2 = 10^{2n+2} + 2\cdot 10^{n+1} + 1.$$

By inspection, we see that the given expression is

$$3^2 [N(n)]^2 + 4^2 [N(n)]^2,$$

which equals $5^2 [N(n)]^2$. As the product of two perfect squares this is itself a perfect square.

Answer 4

We can represent those numbers like this:

$300....060....03 = 3 \cdot 10^{2n+2}+6 \cdot 10 ^{n+1} +3 $

$ 40....080...04 = 4 \cdot 10^{2n+2}+8 \cdot 10 ^{n+1} +4 $ where number of zeros between two numbers is n.

If we put $10^{n+1}=x $

we have $3 \cdot 10^{2n+2}+6 \cdot 10 ^{n+1} +3 = 3x^2+6x+3 = 3(x+1)^2 $

and $4 \cdot 10^{2n+2}+8 \cdot 10 ^{n+1} +4 = 4x^2+8x+4 = 4(x+1)^2 $

Your sum of squares now looks like $[3(x+1)^2]^2+[4(x+1)]^2= 9(x+1)^4+16(x+1)^4=25(x+1)^4=[5(x+1)^2]^2$ which is a perfect square.

Answer 5

Hint $\ $ By $\,\rm\color{#c00}{DS} = $ Difference of Squares we see it's just a scaling of $\ \color{#0a0}{3^2 + 4^2 = 5^2}$

$\qquad\ (5x^2\!+\!10x\!+\!5)^2 - (4x^2\!+\!8x\!+\!4)^2 \overset{\rm\color{#c00}{DS}}= (9x^2\!+\!18x\!+\!9)(x^2\!+\!2x\!+\!1) = (3(x^2\!+\!2x\!+\!1))^2$

$\qquad {\rm i.e.}\ \ \ \ (5(x+1)^2)^2 - (4(x+1)^2)^2 \overset{\rm\color{#c00}{DS}}= 9(x+1)^2\,(x+1)^2 = (3(x+1)^2)^2\qquad\qquad\quad\ $

$\qquad\qquad\qquad\qquad\quad\ \ (5X)^2 - (4X)^2 \overset{\rm\color{#c00}{DS}}= (9X) X\, =\, (3X)^2$

$\qquad\qquad\qquad\qquad\qquad\qquad\ (\color{#0a0}{5^2 - 4^2} \overset{\rm\color{#c00}{DS}}=\, \color{#0a0}{3^2})\,X^2$