# no positive integers

• Apr 28th 2009, 08:42 AM
Fayyaz
no positive integers
Show no positive integers m for which m^4 + 2m^3 + 2m^2 + 2m + 1 is a perfect square. Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?

If so, find all such m.
• Apr 28th 2009, 09:05 AM
running-gag
Quote:

Originally Posted by Fayyaz
Show no positive integers m for which m^4 + 2m^3 + 2m^2 + 2m + 1 is a perfect square. Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?

If so, find all such m.

Hi

$\displaystyle m^4 + 2m^3 + 2m^2 + 2m + 1 = m^4 + 2m^2 + 1 + 2m^3 + 2m = (m^2 + 1)^2 + 2m\:(m^2 + 1)$

$\displaystyle m^4 + 2m^3 + 2m^2 + 2m + 1 = (m^2 + 1)\:(m^2 + 2m + 1) = (m^2 + 1)\:(m+1)^2$
• Apr 28th 2009, 10:50 AM
Soroban
Hello, Fayyaz!

running-gag has an excellent proof, but omitted the punchline.

Quote:

$\displaystyle N \;=\;m^4 + 2m^3 + 2m^2 + 2m + 1 \;=\; m^4 + 2m^2 + 1 + 2m^3 + 2m$

. . $\displaystyle =\; (m^2 + 1)^2 + 2m\,(m^2 + 1) \;=\; (m^2 + 1)\,(m^2 + 2m + 1) \;=\; (m^2 + 1)\,(m+1)^2$

If $\displaystyle N$ is a square, then $\displaystyle (m^2+1)$ must be a square.

. . Hence, $\displaystyle m^2 + 1 \:=\:a^2\quad\hdots$ for some integer $\displaystyle a.$

Then: .$\displaystyle a^2-m^2 \:=\:1 \quad\hdots$ two squares differ by 1.

. . But the only two squares that differ by 1 are: .$\displaystyle a = \pm1,\text{ and }\, m = 0$

Since $\displaystyle m$ must be a positive integer, no solution exists.

• May 3rd 2009, 03:05 AM
Fayyaz
thanks guys

how should i proceed with the next bit:

Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?
• May 5th 2009, 12:41 PM
Hello Fayyaz
Quote:

Originally Posted by Fayyaz
thanks guys

how should i proceed with the next bit:

Are there any positive integers m for which m^4 +m^3 +m^2 +m+1 is a perfect square?

When $\displaystyle m = 3, m^4 +m^3 +m^2 +m+1=121 = 11^2$. So there's certainly one value. Whether there are any more, I don't know.