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Prove:

(A) sum of two squares of two odd integers cannot be a perfect square

(B) the product of four consecutive integers is $1$ less than a perfect square

For (A) I let the two odd integers be $2a + 1$ and $2b + 1$ for any integers $a$ and $b$. After completing the expansion for sum of their squares , I could not establish the link to it not being a perfect t square

For (B) I got stuck at expanding $(a)(a + 1)(a+2) (a + 3)$

Could someone help please ?

Thanks

user1176409
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NetUser5y62
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    Your sum is $4a^2 +4a+4b^2+4b+2$. This is divisible by $2$ but not by $4$, so cannot be a perfect square. – André Nicolas Jul 09 '15 at 04:47
  • For some neat answers for Part B, refer to http://math.stackexchange.com/questions/532737/proving-any-product-of-four-consecutive-integers-is-one-less-than-a-perfect-squa – Shailesh Jul 09 '15 at 06:10
  • @GuoJinLong Where did you get stuck in expanding $a(a+1)(a+2)(a+3)$? Was there nothing you were able to expand? – Erick Wong Jul 09 '15 at 07:04

2 Answers2

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A) \begin{align*} &\,(2a+1)^2+(2b+1)^2=(4a^2+4a+1)+(4b^2+4b+1)=4(a^2+b^2)+4(a+b)+2\\ =&\,2[2(a^2+b^2)+2(a+b)+1]. \end{align*} Since $2(a^2+b^2)$ and $2(a+b)$ are both even, the expression between the brackets is odd because of the $+1$ term. Now, the double of an odd number can never be a perfect square. (Try proving this last statement.)

B) \begin{align*} &\,a(a+1)(a+2)(a+3)=a(a+3)[(a+1)(a+2)]=a(a+3)[a^2+3a+2]\\ =&\,a(a+3)[a(a+3)+2]=a^2(a+3)^2+2a(a+3)=[a^2(a+3)^2+2a(a+3)+1]-1\\ =&\,[a(a+3)+1]^2-1. \end{align*}

triple_sec
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Hint

For part A), it may be helpful to first prove the following:

$n^2 \equiv 0~\text{or}~1\mod{4}$ for all integers $n$.

If $n$ is even, $n=2k$ and $n^2=4k^2\equiv 0\mod 4$. If $n$ is odd, $n=2k+1$ and $n^2=4k^2+4k+1=4(k^2+k)+1\equiv 1\mod 4$. Now, what can you say about the sum of two odd squares modulo 4?

JMoravitz
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