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Show that there exist an infinite number of integers $a^2$ which can also be expressed as the sum $4b+1$.

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    $(2n+1)^2$ ... do the algebra – 2017-02-18
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    Note that odd squares are all of the form $8b+1$ because $n(n+1)$ is even. – 2017-02-18

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Let $a$ be any odd positive integer. Then, $a$ can be expressed in the form, $$a=2k-1$$ where $k\in\mathbb Z^+$.

Then, $$a^2=(2k-1)^2=4k^2-4k+1=4(k^2-k)+1=4b+1$$ where $b=k^2-k\in\mathbb Z^+\cup\{0\}$.

As there are an infinite number of odd positive integers, each of which can be expressed in this form, the statement is proved.

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    Nitpick: Not all odd integers generate distinct values of $a^2$, so you are implicitly using the fact that the mapping $a \mapsto a^2$ is finite-to-one, to go from "infinite number of odd integers" to "infinite number of integers $a^2$". – 2017-02-18
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    @ErickWong Okay, so I simplified it b restricting to positive integers. Thank you for pointing it out. – 2017-02-18